based on maharashtra board.

2. 2.1. 34.2 g of glucose is dissolved in 400 g of water. Calculate
percentage by mass of glucose solution. (7.87 w/w) .
Given- mass solute=34.2g, mass of solution= mass of solvent + solute= 400+34.2= 434.2
% by mass of solute(glucose)
= mass of solute x 100 = 7.87%
Mass of solution

3. 2.2. A solution is prepared by dissolving certain amount of solute in
500 g of water. The percentage by mass of a solute in solution is 2.38.
Calculate mass of solute. (12.19 g). [2 Marks]

4. 3) 4.6 cm3 of methyl alcohol is dissolved in 25.2 g of water.
Calculate: i)% by mass of methyl alcohol. ii)mole fraction of
methyl alcohol and water. (Given density of methyl alcohol
= 0.7952 gcm-3 and C = 12 ,H =1 ,O = 16). [2 Marks]

5. 3) 4.6 cm3 of methyl alcohol is dissolved in 25.2 g of water.
Calculate: i)% by mass of methyl alcohol. ii)mole fraction of
methyl alcohol and water. (Given density of methyl alcohol
= 0.7952 gcm-3 and C = 12 ,H =1 ,O = 16). [2 Marks]

6. 4)12.8 cm3 of benzene is dissolved in 16.8 cm3 of xylene.
Calculate % by volume of benzene (43.24). [2Marks]

7. 5) Calculate mole fraction of HCl in solution of HCl
containing 24.8% of HCl by mass. (H = 1, Cl =35.5,)
(0.1398). 2Marks

8. 6)Calculate the mole fraction of solute in its 2
molal aqueous solution. 2Marks

9. 7)Calculate mole fractions, molality and molarity of HNO3
in a solution containing 12.2% HNO3.
Given density of HNO3 = 1.038 g cm-3 , H=1, N=14, O=16
(xHNO3 = 0.0396 m= 2.205 mol Kg-1 M =2.01 mol dm -3).

11. 8)Sulphuric acid is 95.8% by mass. Calculate mole fraction
and molarity of H2SO4 of density 1.91 cm-3. (H = 1, S = 32,
O = 16) ( X H2SO4 = 0.80730, 17.95M). 2 Marks

12. 8)Sulphuric acid is 95.8% by mass. Calculate mole fraction
and molarity of H2SO4 of density 1.91 cm-3. (H = 1, S = 32,
O = 16) ( X H2SO4 = 0.80730, 17.95M). 2 Marks

13. 9)Aqueous solution of NaOH is marked 10% (w/w). The
density of the solution is 1.070g cm-3 . Calculate
i) molarity, ii)molality, iii)mole fraction of NaOH & water.
Na = 23, H = 10, O = 16, (M = 2.675, m = 2.77, X NaOH =
0.0476, X H2O = 0.9523). 3 Marks

14. 9)Aqueous solution of NaOH is marked 10% (w/w). The
density of the solution is 1.070g cm-3 . Calculate
i) molarity, ii)molality, iii)mole fraction of NaOH & water.
Na = 23, H = 10, O = 16, (M = 2.675, m = 2.77, X NaOH =
0.0476, X H2O = 0.9523). 3 Marks

15. 10)Battery acid is 4.22M aqueous H2SO4 solution, and
has density of 1.21 g cm-3 . What is the molality of H2SO4
? h = 1, S = 32, O = 16 (m = 5.298 mol kg-1). 2Marks

16. 11)The vapour pressure of 2.1% solution of a non-
electrolyte in water at 100 C is 755 mm Hg. Calculate the
molar mass of the solute (58.69 g mol-1). 2Marks

17. 12)The vapour pressure of water at 20 C is 17 mm Hg.
Calculate the vapour pressure of a soluttion containing 2.8
g of urea (NH2CONH2) in 50 g of water N = 14, O = 16, H = 1
(16.71 mm Hg). 2Marks

18. 13)In an expirement,18.04 g of mannitol were dissolved in
100 g of water was lowered by 0.309 mm Hg. From 17.535
mm Hg. Calculate the molar mass of mannitol. (184.3 g
mol-1). 2 Marks

19. 14)A solution containing 0.5126 g of naphthalene (molar mass =
128.17 mol-1) in 50.00 g of CCl4 gives a boiling point elevation of 0.402
K. While a solution of 0.6216 g of unknown solute in the same mass of
the solvent gives a boiling point elevation of 0.6216 K. Find the molar
mass of the unknown solute. (Kb for CCl4 = 5.03 K kg mol-1 of solvent)

20. 15)Calculate the mass in grams of an impurity of molar
mass 100 g mol-1. Which would be required to rise the
boiling point of 50 g chloroform by 0.30 K. (Kb for
chloroform = 3.63 K kg mol-1). 2 Marks

21. 16)Boiling point of a solvent is 80.2 C. When 0.419 g of the
solute molar mass 252.4 g mol-1 was dissolved in 75 g of the
solvent, the boiling point of the solution was found to be 80.256
C. Find the molal elevation constant. (2.53 K kg mol-1).

22. 17)0.6 ml of glacial acetic acid with density 1.06 gm L-1 is
dissolved in 1 kg water and the solution froze at -0.0205 ⁰C.
Calculate vant Hoff factor and Kf for water is 1.86 K kgmol-1.
(i= 1.041, Ka = 1.86 X10-5). [2 Marks]

23. checking

24. 18)Henry’s law constant for solubility of methane in
benzene is 4.27 X 10-5 mm-1 Hg mol dm-3 at constant
temperature. Calculate solubility of methane at 760 mm Hg
pressure at same temperature. [2 marks]
Answer checking
178x10^5