The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125 . The molality of the solution is about

Question

The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125 . The molality of the solution is about (A) 0.70 (B) 0.50 (C) 0.90 (D) 0.80

Tardigrade Admin · Accepted Answer

Relative lowering of vapour pressure = mole fraction of solute (Raoults law)

\frac{P - P _{S}}{P} = X_{2}

\frac{P - P _{S}}{P} = \frac{wM}{mW}

where w = wt. of solute M = mol. wt. of solvent m = mol. wt of solute W = wt. of solvent

0.0125 = \frac{wM}{mW}

Or,

\frac{w}{mW} = \frac{0.0125}{18} = 0.00070

Hence, molality

= \frac{w}{mW} \times 1000

= 0.0007 \times 1000 = 0.70

Q. The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is $0.0125$ . The molality of the solution is about

$0.70$

$0.50$

$0.90$

$0.80$

$0.60$

Q. The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125 . The molality of the solution is about

0.70

0.50

0.90

0.80