\u092c\u093f\u0939\u093e\u0930 \u092c\u094b\u0930\u094d\u0921 \u0907\u0902\u091f\u0930 \u092a\u0930\u0940\u0915\u094d\u0937\u093e 2022 \u0915\u0947 \u0938\u092d\u0940 \u0935\u093f\u0926\u094d\u092f\u093e\u0930\u094d\u0925\u0940 \u0915\u0947 \u0938\u092d\u0940 \u0935\u093f\u0937\u092f \u0915\u0940 \u0938\u092d\u0940 \u092a\u094d\u0930\u0915\u093e\u0930 \u0915\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0915\u093e \u092a\u094d\u0930\u093e\u0930\u0942\u092a \u0914\u0930 PDF \u0935\u0930\u094d\u0917 \u0928\u094b\u091f \u0935\u093f\u0937\u092f\u0935\u093e\u0930 \u0938\u092d\u0940 \u092a\u094d\u0930\u0915\u093e\u0930 \u0915\u0947 study note ( MCQ , Short question long question )

Question 1.<\/span>
\nThe decomposition of H_{2<\/sub>O_{2<\/sub> was studies by titrating is at different intervals of time with KMnO_{4 <\/sub>Calculate the velocity constant for it from the following data, if the reaction is of the first order.<\/span><\/p>\n}}}

$\"\"$ <\/span><\/p>\n

Solution:<\/span>
\nThe first order rate expression is :<\/span>
\nK = \$\\frac{2 \\cdot 303}{t} \\log \\frac{a}{a-x}\$<\/span><\/p>\n

The amount of KMnO_{4<\/sub> is proportional to the amount of H_{2<\/sub>O_{2<\/sub> present., so the volume of KMnO_{4<\/sub> used at zero time corresponds to initial concentration (a) and the volume used after time, corresponds to (a-x) at that time. Inserting these values in the above equation, we get-<\/span><\/p>\n}}}}

K_{600<\/sub> = \$\\frac{2 \\cdot 303}{600} \\log \\frac{22 \\cdot 3}{13 \\cdot 8}\$<\/span>
\n= 0.000837sec^{-1<\/sup><\/span>
\nK_{1200<\/sub> = \$\\frac{2 \\cdot 303}{1200} \\log \\frac{22 \\cdot 8}{8 \\cdot 2}\$<\/span>
\n= 0.000852 sec^{-1<\/sup><\/span><\/p>\n}}}}

The average value of velocity constant.<\/span>
\nK = \$\\frac{0.000837+0.000852}{2}\$<\/span>
\n= 0.000844 sec^{-1<\/sup><\/span><\/p>\n}

Question 2.<\/span>
\nThe optical reaction of sucrose in presence of dil HCl at various intervals is given in the following table:<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

Show that the reaction is of the first order.<\/span>
\nSolution:<\/span>
\nThe inversion of sucrose will be a first order reaction if the above date confirms to the equation:<\/span>
\nK = \$\\frac{2 \\cdot 303}{t} \\log \\frac{a}{(a-x)}\$<\/span>
\n\$=\\frac{2 \\cdot 303}{t} \\log \\left(\\frac{r_{0}-r_{\\infty}}{n-r_{\\infty}}\\right)\$<\/span>
\nWhere, r_{0<\/sub>, r_{1<\/sub> and r_{\u221e<\/sub> are the optical rotation and the start of the reaction, after t and at the completion of rerion respectively.<\/span><\/p>\n}}}

Inserting the values in the above equation, we get-<\/span><\/p>\n

K_{10<\/sub> = \$\\frac{2 \\cdot 303}{10} \\log \\frac{32 \\cdot 4-(-14 \\cdot 1)}{28 \\cdot 8-(-14 \\cdot 1)}\$<\/span>
\n\$=\\frac{2 \\cdot 303}{10} \\log \\frac{46 \\cdot 5}{42 \\cdot 9}\$<\/span>
\n= 0.008060min^{-1<\/sup><\/span><\/p>\n}}

K_{20<\/sub> = \$\\frac{2 \\cdot 303}{20} \\log \\frac{32 \\cdot 4-(-14 \\cdot 1)}{25 \\cdot 5-(-14 \\cdot 1)}\$<\/span>
\n\$=\\frac{2 \\cdot 303}{20} \\log \\frac{46 \\cdot 5}{39 \\cdot 9}\$<\/span>
\n= 0.008037min^{-1<\/sup><\/span><\/p>\n}}

K_{40<\/sub> = \$\\frac{2 \\cdot 303}{40} \\log \\frac{32 \\cdot 4-(-14 \\cdot 1)}{19 \\cdot 6-(-14 \\cdot 1)}\$<\/span>
\n\$=\\frac{2 \\cdot 303}{40} \\log \\frac{46 \\cdot 5}{33 \\cdot 7}\$<\/span>
\n= 0.008054min^{-1<\/sup><\/span><\/p>\n}}

K_{100<\/sub> = \$\\frac{2 \\cdot 303}{100} \\log \\frac{32 \\cdot 4-(-14 \\cdot 1)}{6 \\cdot 7-(-14 \\cdot 1)}\$<\/span>
\n\$=\\frac{2 \\cdot 303}{100} \\log \\frac{46 \\cdot 5}{20 \\cdot 8}\$<\/span>
\n= 0.008054min^{-1<\/sup><\/span>
\nThe constancy in the value of K shows that the reaction is of first order.<\/span><\/p>\n}}

Question 3.<\/span>
\nDecomposition of diazobenzene chloride was followed at constant temperature by measuring the volume of nitrogen evolved at different times.<\/span>
\nThe date is given below<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

Calculate the specific reaction rate and order of reaction<\/span>
\nSolution:<\/span>
\nDiazobenzene chloride decomposition as:<\/span>
\nC_{6<\/sub>H_{5<\/sub> \u2013 N = N-Cl \u2192 C_{6<\/sub>H_{5<\/sub>Cl + N_{2<\/sub><\/span>
\nThus the amount of N_{2<\/sub> evolved will be a measure ofdiazobenzene chloride decomposed i.e.x. The total N_{23<\/sub> evolved at infinite time will thus give the initial concentration i.e. a. Now substituting the values in the first order rate<\/span>
\nexpression,<\/span><\/p>\n}}}}}}}

$\"\"$ <\/span><\/p>\n

The constancy in the value of K shows that the reaction \u00a1s of the first order. The specific rate constant K is thus given by:<\/span>
\nK = \$\\frac{0.00322+0 \\cdot 00336+0 \\cdot 00326}{3}\$<\/span>
\n= 0.00328min^{-1<\/sup><\/span><\/p>\n}

12th Chemistry Numericals Important Questions with Solutions<\/span><\/p>\n

Question 4.<\/span>
\nA first order reaction is 15% complete in 20 minutes. How long will it take to be 60% complete?<\/span>
\nSolution:<\/span>
\nFor a first order reaction, we have-<\/span>
\nK = \$\\frac{2 \\cdot 303}{t} \\log _{10} \\frac{a}{(a-x)}\$<\/span>
\nor,<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

Here x_{1<\/sub> =\$\\frac{15}{100}\$ a_{1<\/sub><\/span>
\n= 0.15a_{1<\/sub>,t_{1<\/sub> = 20<\/span>
\nx_{2<\/sub> = \$\\frac{60}{100}\$a_{2<\/sub><\/span>
\n= 0.6a_{2<\/sub>,t_{2<\/sub> = ?<\/span>
\nInserting these values in equation (1), we get-<\/span><\/p>\n}}}}}}}}

$\"\"$ <\/span><\/p>\n

Question 5.<\/span>
\nGiven that the temperature co-efficient for the saponification of ethyl acetate by NaOH is 1.75. Calculate the activation energy.<\/span>
\nAnswer:<\/span>
\nGiven that-<\/span>
\n\$\\frac{K_{2}}{K_{1}}\$ = 1.75 T_{1<\/sub>= (25 +273) = 298 K T_{2<\/sub> = (35 + 273) = 308 K<\/span>
\n(Since temperature co-efficient is ration of rate constant at 35\u00b0 and 25\u00b0C respectively).<\/span><\/p>\n}}

$\"\"$ <\/span><\/p>\n

E_{a<\/sub> = \$\\frac{2.303 \\times 298 \\times 308 \\times 1.987}{1}\$ log 1.75 cal mol^{-1<\/sup><\/span>
\nor,<\/span>
\nE_{a<\/sub> = 10.207k cal mol^{-1<\/sup><\/span><\/p>\n}}}}

Colligative Properties of Solutions<\/span>
\nOsmotic Pressure<\/span><\/p>\n

Question 6.<\/span>
\nCalculate the value of R in litre-atomosphere, if the osmotic pressure of a solution containing 45g. of sucrose dissolved per liter of solution at 2\u00b0 C is 3.0 atmospheres.<\/span>
\nSolution:<\/span>
\nThe mol wt. (m) of sucrose (C_{12<\/sub>H_{22<\/sub>O_{11<\/sub>)<\/span>
\n= 12 \u00d7 12 + 1 \u00d7 22 + 16 \u00d7 11<\/span>
\n= 342<\/span>
\nT = 2 + 273<\/span>
\n= 275 K<\/span><\/p>\n}}}

We known that p = \$\\frac{w R T}{m v}\$<\/span>
\nor,<\/span>
\nR = \$\\frac{P . m \\cdot V}{w T}\$<\/span>
\nR = \$\\frac{3 \\times 342 \\times 1}{45 \\times 275}\$<\/span>
\n= 0.0829 lit-atm.<\/span><\/p>\n

12th Chemistry Numericals Important Questions with Solutions<\/span><\/p>\n

Question 7.<\/span>
\nCalculate the osmotic pressure of 5% solution of glucose C_{6<\/sub> H_{12<\/sub>O_{6<\/sub> at 18\u00b0 C.<\/span>
\nSolution:<\/span>
\nGiven that, T = ( 18+273) = 291 K, w = 5g<\/span>
\nV = 100ml = \$\\frac{1}{10}\$ = 0.1L and mg glucose = 180<\/span>
\nwe know that P = \$\\frac{\\boldsymbol{w}}{\\mathrm{m} \\cdot \\boldsymbol{V}}\$RT<\/span>
\nor,<\/span>
\np = \$\\frac{5}{180 \\times 0 \\cdot 1}\$ \u00d7 0.0821 \u00d7 291<\/span>
\n= 6.636 atmosphere.<\/span><\/p>\n}}}

Question 8.<\/span>
\nA solution containing 8 g of a substance per 100 ml was found to have an osmotic pressure of 500 cms of Hg at 27\u00b0C. Calculate the molecular weight of the substance.<\/span>
\nSolution:<\/span>
\nGiven that, w = 8g, V= 100 ml = 0.1 L<\/span>
\n\\frac{500}{76}P = 500 cm = \$\\frac{500}{76}\$ atmosphere<\/span>
\nand T = (27 + 273) = 300 K<\/span>
\nWe know that, p = \$\\frac{w}{m V}\$RT<\/span>
\nor,<\/span>
\nm = \$\\frac{w R T}{p . V}\$<\/span>
\n\$=\\frac{2 \\times 0.0821 \\times 300 \\times 76}{500 \\times 0.1}\$<\/span>
\n= 299.4<\/span><\/p>\n

Question 9.<\/span>
\nA 1.02 % solution of glycerine is isotonic with 2% solution of glucose. What is the molecular weight of glycerine?<\/span>
\nSolution:<\/span>
\nFor isotonic solution we have<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

or,<\/span>
\n\$\\frac{1 \\cdot 02}{m_{1} \\times 0.1}=\\frac{2}{180 \\times 0.1}\$ [\u2235 Mol. wt. of glucose = 180]<\/span>
\n\u2234 m_{1<\/sub> = \$\\frac{180 \\times 0.1 \\times 1.02}{2 \\times 0.1}\$<\/span>
\n= 91.8<\/span><\/p>\n}

Question 10.<\/span>
\nA solution containing 8.6 g\/litre of urea (mol. wt. = 60) was found to be isotonic with a 5% solution of an organic solute. Calculate the molecular weight of the solute.<\/span>
\nSolution:<\/span>
\nFor isotomic solution, we have-<\/span><\/p>\n

$\"12th$ <\/span><\/p>\n

or,<\/span>
\n\$\\frac{8.6}{60 \\times 1}=\\frac{5}{m_{2} \\times 0.1}\$<\/span>
\nor,<\/span>
\nm_{2 <\/sub> = \$\\frac{60 \\times 5 \\times 1}{8.6 \\times 0.1}\$ = 348.9<\/span><\/p>\n}

Question 11.<\/span>
\n10 g of a substance dissolved in 100 ml of water gave rise to an osmotic pressure 500 cm of Hg at 27\u00b0C. Calculate the molecular weight of the substance.<\/span>
\nSolution- We know that-<\/span>
\nm = \$\\frac{w R T}{P V}\$<\/span>
\nGiven that, w = 4.0 ; P = \$\\frac{500}{76}\$atm; V = 0.1 litre<\/span>
\nR = 0.082 1 and T=(27+273) = 300 K<\/span>
\n\u2234 m = \$\\frac{4 \\times 0.082 \\times 300 \\times 76}{500 \\times 0.1}\$ = 149.5<\/span><\/p>\n

Question 12.<\/span>
\nCalculate the osmotic pressure of \$\\frac{\\boldsymbol{M}}{\\mathbf{1 0}}\$ solution of cane sugar at 27\u00b0 C.<\/span>
\nSolution:<\/span>
\nMolecular wt. of cane sugar (C_{12<\/sub> H_{22<\/sub> O_{11<\/sub>) = 342<\/span>
\nFor \$\\frac{\\boldsymbol{M}}{\\mathbf{1 0}}\$ solution, it means that \$\\frac{1}{10}\$ g mole of cane sugar is dissolved in 1 litre of solution i.e. =\$\\frac{342}{10}\$g = 34.2 of cane sugar dissolved in 1 litre.<\/span>
\n\u2234 W = 34.2g,V = 1 litre.<\/span>
\nT = 27+273 = 300K<\/span>
\nand m = 342<\/span>
\nwe know that P = \$\\frac{w R T}{m V}\$<\/span>
\n\$=\\frac{34.2 \\times 0.0821 \\times 300}{342 \\times 1}\$<\/span>
\nor,<\/span>
\nP = 2.463 atmosphere.<\/span><\/p>\n}}}

Question 13.<\/span>
\nCalculate the amount of urea dissolved per litre if its equeour Solution is isotonic with 10% cane sugar. Molecular weight of urea = 60.<\/span>
\nSolution:<\/span>
\nFor urea = m_{1<\/sub> = 60; V_{1<\/sub> = 1 litre, w_{1<\/sub> = ?<\/span>
\nFor cane sugar = w_{2<\/sub> = 10 g; m_{2<\/sub> = 342, V_{2<\/sub>= 0.1 lit.<\/span>
\nFor isotonic solution:<\/span>
\n\$\\frac{w_{1}}{m_{1} V_{1}}=\\frac{m_{2}}{m_{2} V_{2}}\$<\/span>
\nor,<\/span>
\n\$\\frac{w_{1}}{60 \\times 1}=\\frac{10}{342 \\times 0.1}\$<\/span>
\nor,<\/span>
\nw_{1<\/sub> = \$\\frac{10 \\times 60 \\times 1}{342 \\times 0.1}\$<\/span>
\n= 17.6 g\/litre<\/span><\/p>\n}}}}}}}

Question 14.<\/span>
\nAt 298 K, 100 ml of a solution containing 3.002g of an unidentified solute exhibts an osmotic pressure of 2.25 atm. What is the molecular mass of the solute.<\/span>
\nSolution:<\/span>
\nGiven that V = 100 ml = 0.1 litre, w = 3.002g, m = ?<\/span>
\nT = 298 K,P = 2.25 atm and R = 0.0821<\/span>
\nWe know that<\/span>
\nm = \$\\frac{w R T}{P V}\$ = \$\\frac{3.002 \\times 0.0821 \\times 298}{2.25 \\times 0.1}\$ = 288.6<\/span><\/p>\n

12th Chemistry Numericals Important Questions with Solutions<\/span><\/p>\n

Lowering of Vapour Pressure<\/span><\/p>\n

Question 15.<\/span>
\nA current of dry air was drawn through solution of 3.5 g of a ubstance in 100g of ethyl alcohol and then passed through ethyl alcohol lone. The loss of weight of the solution was 0.8759 g and of the alcohol 0241g.Calculate the molecular weight of dissolved substance.<\/span>
\nSolution:<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

\$=\\frac{0.0241}{0.8759+0.0241}\$<\/span>
\n\$=\\frac{0.241}{0.9000}\$<\/span>
\nAccording to Rault\u2019s Law for dilute solutions, we get-<\/span>
\n\$\\frac{P-P_{s}}{P}=\\frac{n}{N}=\\frac{w M}{w M}\$<\/span>
\nw = 3.5, M = Mol. wt. of C_{2<\/sub> H_{5<\/sub> OH = 46. W = 100, m = ?<\/span>
\nSubstituting these values in Rault\u2019s Law eqn., we get<\/span>
\n\$\\frac{0.0241}{0.9000}=\\frac{3.5 \\times 46}{m \\times 100}\$<\/span>
\n\u2234 m = \$\\frac{3.5 \\times 46 \\times 0.9}{100 \\times 0.0241}\$<\/span>
\n\u2234 The Mol. wt. of the dissolved substance = 60.13.<\/span><\/p>\n}}

Question 16.<\/span>
\nAt 20\u00b0C the vapour pressure of ether is 442 m.m. when 6.1g of Benzoic acid are dissolved in 50g of ether, the vapour pressure falls to 410 m.m. What is the molecular weight of Benzoic acid?<\/span>
\nSolution:<\/span>
\nAccording to Rault\u2019s law-<\/span>
\n\$\\frac{P-P_{S}}{P}=\\frac{n}{N}\$ (For dilute solutions)<\/span>
\n\$=\\frac{w M}{W m}\$<\/span>
\nwhere w = wt. of solute = 6.1 : p = 442 m.m.<\/span>
\nW = wt. of solvent = 50 : P_{s<\/sub> = 410 m.m.<\/span>
\nM = mol. wt. of solvent = 74 : P = P_{s<\/sub> = 442 \u2013 410<\/span>
\nm = mol. wt. of solute = ?<\/span>
\nSubtituting these values in the eqn. we have-<\/span>
\n\$\\frac{32}{442}=\\frac{6.1 \\times 74}{m \\times 50}\$<\/span>
\nor,<\/span>
\nm = 124.7<\/span><\/p>\n}}

Question 17.<\/span>
\nDry air was drawn in succession thought a series of bulbs containing 4.27g of a substance X in 52.68g of ethyl alcohol and through similar set of bulbs containing pure alcohol, the indrawn air and the two sets of bulbs were at the same \u2018constant temperature. The. loss of weight in first series of bulbs was 1.292g. and in second series .0313g. Calculate the mole wt.of X.<\/span>
\nSolution:<\/span><\/p>\n

$\"\"$ <\/span><\/p>\n

\$=\\frac{.0313}{0.0313+1.292}\$<\/span>
\n\$=\\frac{0.0313}{1.3233}\$<\/span>
\nAccording to Rault\u2019s Law eqn :<\/span>
\n\$\\frac{P-P_{S}}{P}=\\frac{w M}{W m}\$<\/span>
\nw = 4.257 g,W = 52.68g, M = 46(mol. wt. of C_{2<\/sub>H_{5<\/sub>OH)<\/span>
\nSubstituting these values in the above eqn.<\/span>
\n\$\\frac{0.0313}{1.3233}=\\frac{4.257 \\times 46}{m \\times 52.68}\$<\/span>
\n\u2234 m = \$\\frac{4.257 \\times 46 \\times 1.3233}{0.0313 \\times 52.68}\$<\/span>
\n= 157.1<\/span><\/p>\n}}

Question 18.<\/span>
\nAir was drawn through a solution containing 38g of solute in 100g of water and then through water. The loss of weight of water was .0551 g and the total weight of water absorbed in sulphuric acid tube was 2.2117g. Find the molecular weight of the dissolved substance.<\/span>
\nSolution:<\/span>
\n\$\\frac{P-P_{s}}{P}=\\frac{w M}{m W}\$<\/span>
\n\$\\frac{0.0551}{2.2117}=\\frac{38 \\times 18}{m \\times 100}\$<\/span>
\n\u2234 m = \$\\frac{38 \\times 18 \\times 2.2117}{.055 \\times 100}\$<\/span>
\n= 274.5<\/span><\/p>\n

Question 19.<\/span>
\nDry air was passed through a solution containing 50g of a solute in 100g of water and then through water. The loss in weight of water is 0.05g. The wet air then passed through sulphuric acid whose weight increased by 2.00g. What is the molecular weigh t of the dissolved substance.<\/span>
\nSolution:<\/span>
\np \u2013 p_{s<\/sub> = \$\\frac{w M}{m V} \\cdot \\frac{.05}{2.00}\$<\/span>
\n\$=\\frac{50 \\times 18}{m \\times 100}\$<\/span>
\n\u2234 \$=\\frac{50 \\times 18 \\times 2.00}{0.5 \\times 100}\$<\/span>
\n= 360<\/span><\/p>\n}

12th Chemistry Numericals Important Questions with Solutions<\/span><\/p>\n

Elevation of Boiling Point<\/span><\/p>\n

Question 20.<\/span>
\nThe boiling point of a solution Of 3.40 gms. of BaCl_{2<\/sub> (Mol. wt. 208,4) in 100 gms. of water is 100.208\u00b0C. Calculate the degree of dissociation BaCl_{2<\/sub> if the molecular elevation for water is 5.2.<\/span>
\nSolution:<\/span>
\nCalculation of observed mol. wt.<\/span>
\nm = \$\\frac{100 \\times K \\times w}{\\Delta T \\times W}\$<\/span>
\n\$=\\frac{100 \\times 5.2 \\times 3.4}{280 \\times 100}\$<\/span>
\n= 85.0<\/span>
\nBut calculated mol.wt. = 208.4<\/span><\/p>\n}}

$\"\"$ <\/span><\/p>\n

Particles after dissociation<\/span>
\n1 \u2013 x + x + x + x = 1 + 2x<\/span>
\nBut<\/span>
\n\$\\frac{\\text { Normal mol. wt. }}{\\text { observed mol. wt. }}=\\frac{1+2 x}{1}\$<\/span>
\n\$=\\frac{208.4}{85}\$<\/span>
\n\u2234 x = 7259<\/span>
\n\u2234 Degree of dissociation of BaCl_{2<\/sub> = 72.59%<\/span><\/p>\n}

Question 21.<\/span>
\n1.23 g of a substance dissolved in 10g of water raised the boiling point by 0.39\u2033C. Calculate the molecular weight (K = 5.2\u00b0C)<\/span>
\nSolution:<\/span>
\nWe know that:<\/span>
\nm = \$\\frac{100 \\times K \\times w}{\\Delta T \\times W}\$<\/span>
\n\$=\\frac{100 \\times 5.2 \\times 1.23}{10 \\times 3.9}\$<\/span>
\n\u2234 Mol. wt. of the substance =164<\/span><\/p>\n

Question 22.<\/span>
\n1.55 g of urethane dissolved in 34.22g of methyl alcohol raised the boiling point by 0.23\u2033. The elevation constant for methyl alcohol is 8.8. Calculate the molecular weight of urethane.<\/span>
\nSolution:<\/span>
\nAs given in the question, we have-<\/span>
\nw = 1 .55 g \u0394t = 0.32\u00b0<\/span>
\nW = 34.22g = 8.8<\/span>
\nNow m = \$\\frac{100 \\mathrm{Kw}}{\\Delta \\mathrm{TW}}\$ = \$\\frac{100 \\times 8.8 \\times 1.55}{0.32 \\times 34.22}\$ = 124.6<\/span>
\n\u2234 Mol wt. of urethane = 124.6.<\/span><\/p>\n

Question 23.<\/span>
\nIf the boiling point of water is raised by 0.071\u00b0 by dissolving 1.17 g of cane sugar (C_{12<\/sub> H_{22<\/sub>O_{11<\/sub>) in 25g of the solvent, find out the molecular elevation of boiling point<\/span>
\nSolution:<\/span>
\nAs given in the question,<\/span>
\nw = 1.17 gm \u0394T = .071\u00b0<\/span>
\nW = 25 gm m = mol. wt. of cane sugar<\/span>
\n= 144 + 22+ 176 = 342<\/span>
\nNow<\/span>
\nm = \$\\frac{1000 \\mathrm{Kw}}{\\Delta T W}\$<\/span>
\nK = \$\\frac{m \\times \\Delta T \\times W}{100 w}\$<\/span>
\n\$=\\frac{342 \\times 0.071 \\times 25}{.1000 \\times 1.17}\$ = 5.18<\/span>
\nHence the molecular elevation of boiling point = 518<\/span><\/p>\n}}}

12th Chemistry Numericals Important Questions with Solutions<\/span><\/p>\n

Question 24.<\/span>
\nThe boiling point of a solution of 0.105g Of substance in 7.92g of ether was higher than that of the pure solvent by 0.2\u00b0. Find the mol wt. of the solute (K_{b<\/sub> = 21.1)<\/span>
\nSolution:<\/span>
\nWe know that-<\/span>
\nm = \$\\frac{100 \\times K \\times w}{\\Delta T \\times W}\$<\/span>
\n\$=\\frac{100 \\times 21 \\cdot 6 \\times 0 \\cdot 105}{0 \\cdot 2 \\times 7 \\cdot 92}\$<\/span>
\n= 139.8<\/span>
\n\u2234 Mol, wt. of the substance = 139.8<\/span><\/p>\n}

Depression in Freezing Point<\/span><\/p>\n

Question 25.<\/span>
\nCalculate the molar depression constant of water.<\/span>
\nSolution:<\/span>
\nLatent heat of fusion (LJ) = 80.0 Cal\/g<\/span>
\nFreezing point of water (T) = 0\u00b0C = 0 + 273 = 273\u00b0A<\/span>
\n\u2234 K = \$\\frac{R T^{2}}{1000 L_{f}}\$<\/span>
\n\$=\\frac{2 \\times(273)^{2}}{1000+80}\$<\/span>
\n= 1.863\u00b0<\/span>
\n= 1.863\u00b0<\/span><\/p>\n

12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions<\/span>\r\n12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions 12th Chemistry Numericals Important Questions with Solutions<\/span><\/span><\/pre>\n","protected":false},"excerpt":{"rendered":"Bihar Board 12th Chemistry Numericals Important Questions with Solutions in English \u092c\u093f\u0939\u093e\u0930 \u092c\u094b\u0930\u094d\u0921 \u0907\u0902\u091f\u0930 \u092a\u0930\u0940\u0915\u094d\u0937\u093e 2022 \u0915\u0947 \u0938\u092d\u0940 \u0935\u093f\u0926\u094d\u092f\u093e\u0930\u094d\u0925\u0940 \u0915\u0947 \u0938\u092d\u0940 \u0935\u093f\u0937\u092f \u0915\u0940 \u0938\u092d\u0940 \u092a\u094d\u0930\u0915\u093e\u0930 \u0915\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0915\u093e \u092a\u094d\u0930\u093e\u0930\u0942\u092a \u0914\u0930 PDF \u0935\u0930\u094d\u0917 \u0928\u094b\u091f \u0935\u093f\u0937\u092f\u0935\u093e\u0930 \u0938\u092d\u0940 \u092a\u094d\u0930\u0915\u093e\u0930 \u0915\u0947 study note ( MCQ , Short question long question ) Bihar Board\u00a0Class 10 & 12\u00a0Science all subject Note […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"rank_math_lock_modified_date":false,"footnotes":""},"categories":[742,746],"tags":[2468,2467,2471,2474,2476,2472,2475,2473,2461,2462,2466,2465,2477,2460,2463,2464,2470,2469],"class_list":["post-7054","post","type-post","status-publish","format-standard","hentry","category-class-12","category-class-12-chemistry","tag-class-12-chemistry-chapter-2-important-numericals","tag-class-12-chemistry-important-numericals","tag-class-12-chemistry-important-questions","tag-class-12-chemistry-important-questions-2020","tag-class-12-chemistry-important-questions-4ono","tag-class-12-chemistry-important-questions-for-board-exam-2020","tag-class-12-chemistry-important-questions-in-hindi","tag-class-12-chemistry-important-questions-neb","tag-class-12-chemistry-important-questions-pdf","tag-class-12-chemistry-important-questions-with-answers","tag-class-12-chemistry-most-important-questions","tag-class-12-chemistry-ncert-important-questions-pdf","tag-class-12-chemistry-numericals","tag-class-12-chemistry-solutions-chapter-important-questions","tag-class-12-imp-question-chemistry","tag-class-12-important-question-chemistry","tag-important-numericals-of-chemistry-class-12","tag-numerical-problems-in-chemistry-class-12"],"_links":{"self":[{"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/posts\/7054"}],"collection":[{"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/comments?post=7054"}],"version-history":[{"count":0,"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/posts\/7054\/revisions"}],"wp:attachment":[{"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/media?parent=7054"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/categories?post=7054"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/examfriends.com\/wp-json\/wp\/v2\/tags?post=7054"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}