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Std: 11th Standard - Chemistry

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Questions:

Generic 3

4 x 3 = 12

Vapour pressure of a pure liquid A is 10.0 torr at 27°C. The vapour pressure is lowered to 9.0 torr on dissolving one gram of B in 20 g of A. If the molar mass of A is 200 then calculate the molar mass of B.

0.75 g of an unknown substance is dissolved in 200 g water. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5 K Kg mol^-1 then, calculate the molar mass of unknown substance

2g of a non electrolyte solute dissolved in 75 g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5.12 K Kg mol^-1. Find the molar mass of the solute.

The depression in freezing point is 0.24K obtained by dissolving 1g NaCl in 200g water. Calculate van’t-Hoff factor. The molal depression constant is 1.86 K Kg mol^-1.

Generic 5

1 x 5 = 5

The observed depression in freezing point of water for a particular solution is 0.093^o C. Calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol^-1.

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Answers Key:

Generic 3

4 x 3 = 12

\(P_A^o\) = 10 torr, P_solution = 9 torr
W_A = 20 g W_B = 1 g
M_A = 200 g mol^-1 M_B = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

ΔT_b = K_b m
= K_b x W₂ x 1000 / M₂ x W₁
M₂ = K_b x W₂ x 1000 / ΔT_b x W₁
= 7.5 x 0.75 x 1000 / 0.15 x 200
= 187.5 g mol^-1

W₂ = 2g, W₁ = 75 g
ΔT_f = 0.2 K , K_f = 5.12 K Kg mol^–1
M₂ = ?
\(M_2={K_f\times W_2\times 1000\over \Delta T_f\times W_1}={5.12\times 2\times 1000\over 0.2\times 75}\)
= 682.66 g mol^-1.

Molar mass of solute
\(={1000\times K_f\times mass\ of\ NaCl\over \Delta T_f\times mass\ of\ solvent}\)
\(={1000\times 1.86\times 1\over 0.24\times 200}\)
= 38.75 g mol^-1
= 38.75 g mol
Theoretical molar mass of NaCl is
i = \({Theoretical\ molar\ mass\over Experimental\ molar\ mass}={58.5\over 38.75}=1.50\)

Generic 5

1 x 5 = 5

\(\Delta T_f=0.093^oC=0.093K\)
m = ?
K_f = 1.86K Kg mol^-1
\(\Delta T_f=K_f.m\)
\(\therefore m={\Delta T_f\over K_f}\)
\(={0.093K\over 1.86\ K\ Kg\ mol^{-1}}\)
= 0.05 mol Kg^-1
= 0.05 m.