Question

6g. of a non volatile solute is dissolved in 90g,
of water, such that the lowering in vapour
pressure is 2%. The molecular weight of the
solute ​

Answer 1

The molecular weight of the  solute ​ is 58.8 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 : 6 g of non volatile solute is present in 90 g of water

moles of solute = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{6g}{Mg/mol}$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{90g}{18g/mol}=5moles$

Total moles = moles of solute + moles of solvent (water) =$\frac{6g}{Mg/mol}+5$

$x_2$ = mole fraction of solute  =$\frac{\frac{6g}{Mg/mol}}{\frac{6g}{Mg/mol}+5}$

$\frac{2}{100}=1\times \frac{\frac{6g}{Mg/mol}}{\frac{6g}{Mg/mol}+5}$

$M=58.8g/mol$

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Answer 2

Answer:

60gm

Explanation: