Question

A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute.​

Answer 1

Answer:

As follow

Explanation:

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Answer 2

The molar mass of the solute is 179 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,

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

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute   =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 10 g of non-volatile solute is present in 200 g of water

moles of solute = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{10g}{Mg/mol}$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{200g}{18g/mol}=11.1moles$

Total moles = moles of solute + moles of solvent (water) = $\frac{10g}{Mg/mol}+11.1$

$x_2$ = mole fraction of solute  =$\frac{\frac{10g}{Mg/mol}}{\frac{10g}{Mg/mol}+11.1}$

$\frac{32-31.84}{32}=1\times \frac{\frac{10g}{Mg/mol}}{\frac{10g}{Mg/mol}+11.1}$

$M=179g/mol$

Thus the molar mass of the solute is 179 g/mol

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