Question

If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?

• A. 21.3 g/ mol
• B. 23.1 g/mol
• C. 32.1 g/mol
• D. 12.3 g/mol

Answer 1

The correct option is A 21.3 g/ mol

Given,
$\Delta p_1/p_1^0=0.005$
Mass of solute, $m_w=0.5g$
Mass of solvent, $m_S=500g$
Number of moles of solute, $n_S=\frac{0.5}{M_w}$, where $M_w$ is the molecular weight of the solute.
Number of moles of solvent, $n_{Solvent}=\frac{500}{106}=4.7mol$
Since $m_S<<m_{Solvent},$ the solution can be considered to be dilute. Hence, equation for relative lowering for vapor pressure becomes:
$\Delta p_1/p_1^0=n_2/(n_1+n_2)\approx n_2/n_1$
On substituting values,
$0.005=\left(0.5/M_w\right)/4.7$
On solving, $M_w=21.3g/mol.$