Question

When some NaCl was dissolved in water, the freezing
point depression was numerically equal to twice the
molal depression constant. The relative lowering of
vapour pressure of the solution nearly is

(1) 0.036
(2) 0.018
(3) 0.0585
(4) 0.072

Answer 1

the relative lowering of vapour pressure of the solution is nearly 0.036...

hope it help u..

Answer 2

The relative lowering of vapour pressure of the solution nearly is 0.036.

Explanation:

Let $k_f$ be the molal depression constant.

According to question

Depression in freezing point=$\Delta T_f=2K_f$

We know that

$\Delta T_f=mK_f$

Where $\Delta T_f$=Freezing point depression

$K_f$=Molal depression constant

m=Molality of solution

Substitute the values then we get

$2K_f=mK_f$

$m=\frac{2K_f}{K_f}=2$

Molality=$\frac{\frac{W_B}{M_B}}{W_A(in g)}\times 1000$

Where $n_B$=Number of moles of solute

$W_A$=Given mass of solvent (in g)

Relative lowering in vapour pressure of solution=$\frac{\frac{W_B}{M_B}}{\frac{W_A}{M_A}}=\frac{M_A\times m}{1000}$

Where $M_A$=Molar mass of solvent

Molar mass of water$(H_2O)=2(1)+16$

$M_A=18$g

Relative lowering in vapour pressure of solution=$\frac{18\times 2}{1000}=0.036$

Hence, the relative lowering of vapour pressure of the solution nearly is 0.036.

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