Question

On dissolving a non-volatile and non-electrolyte solute in water, the vapour pressure of water decreases to 740 mm hg. the molality of solution approximately will be: (vapour pressure of pure water : 760 mm hg)

Answer 1

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you understand the solution

Answer 2

Answer: 1.48 mol/kg

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,

$\frac{p^o-p_s}{p^o}=x_2$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 760 mmHg

$p_s$ = vapor pressure of solution  = 740 mmHg

$x_2$ = mole fraction of solute = ? g

$\frac{760-740}{760}=x_2$

$x_2=0.026$

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molarity=\frac{n\times 1000}{W_s}$

where,

n= moles of solute  = 0.026

moles of solvent (water) = 1-0.026 =0.973

$W_s$ = weight of solvent in g=$moles\times {\text {molar mass}}=0.973\times 18=17.52g$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.026moles\times 1000}{17.52g}=1.48mole/kg$

Therefore, the molality of solution will be 1.48 mole/kg.