A non-volatile, non-electrolyte solid solute when dissolved in 40 g...

From the relative lowering of vapour pressure, we have:

$\dfrac{P^0 - P_s}{P_s} = \dfrac{n_B}{n_A}$

where $n_B$ is the number of moles of solute and $n_A$ is the number of moles of solvent.

Substituting the given values:

$\dfrac{760 - 750}{750} = \dfrac{n_B}{n_A}$

$\dfrac{10}{750} = \dfrac{n_B}{n_A} \Rightarrow n_A = 75 n_B$

The elevation in boiling point is given by:

$\Delta T_b = T_b - T_b^0 = 320 - 319.5 = 0.5$ K

We also know that $\Delta T_b = K_b \times m$, where $m$ is the molality of the solution.

$m = \dfrac{n_B}{\text{Mass of solvent in kg}} = \dfrac{n_B}{0.04}$

Substituting the values into the elevation in boiling point equation:

$0.5 = 0.3 \times \dfrac{n_B}{0.04}$

$n_B = \dfrac{0.5 \times 0.04}{0.3} = \dfrac{0.02}{0.3} = \dfrac{1}{15}$ mol

Now, substituting $n_B$ back to find the number of moles of the solvent $n_A$:

$n_A = 75 \times \dfrac{1}{15} = 5$ mol

Answer: $5$