Question

28. a) 5.8 g of non-volatile, non-electrolyte was dissolved in 100 g of carbon disulphide (CS2).
The vapour pressure of the solution was found to be 190 mm of Hg. Calculate molar mass
of the solute. Given: vapour pressure of pure CS2 is 195 mm of Hg and molar mass of
CS2 is 76 g mol?
en ideal and non ideal solutions
(3+2)

Answer 1

Answer: The molar mass of solute is 172 g/mol.

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent  $(CS_2)$ = 195 mmHg

$p_s$ = vapor pressure of solution  = 190 mmHg

$w_2$ = mass of solute  = 5.8 g

$w_1$ = mass of solvent   $(CS_2)$=100 g

$M_1$ = molar mass of solvent $(CS_2)$  = 76 g/mole

$M_2$ = molar mass of solute =? g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{195-190}{195}=\frac{5.8\times 76}{100\times M_2}$

$M_2=172g/mol$

Therefore, the molar mass of solute is 172 g/mol.

Answer 2

Answer ⇒ The molar mass of non-volatile solute is 163.096 g/mole.

Explanation ⇒ Vapour Pressure of solution = 190 mm of Hg.

Vapour Pressure of Solvent CS₂ = 195 mm of Hg.

Using the formula,

Relative lowering of vapor pressure of solvent = mole fraction of non-volatile solute.

∴ (195 - 190)/190 = mole fraction of solute.

∴ 5/190 = mole fraction of solute.

∴ 1/38 = mole fraction of solute.

For mole fraction of solute,

Mass of non-volatile solute = 5.8 g.

Molar mass of non-volatile solute = M. [Say.]

Mass of Solvent = 100 g.

Molar mass of solvent = 76 gm.

∴ Mole fraction of solute = $\frac{\frac{5.8}{M} }{\frac{5.8}{M} + \frac{100}{76} }$

Let 5.8/M = a.

∴ Mole fraction = a/(a + 1.32)

∴ 1/38 = a/(a + 1.32)

∴ (a + 1.32) = 38a

∴ a + 1.32 = 38a

∴ 37a = 1.32

∴ a = 0.036

∴ 5.8/M =  0.036

∴ M = 163.096 g/mole.

Hence, the molar mass of non-volatile solute is 163.096 g/mole.

Hope it helps.