Determine the Boiling Point of Diluted Solution M

To determine the boiling point of the diluted solution, we can use the relation between the elevation in boiling point and the molality of the solution.

1. Calculate the Molality of the Solution:
   Let the mole fraction of water (solvent) be $X_{\text{solvent}} = 0.9$.
   Since the solution contains a non-volatile solute, the mole fraction of the solute is:
   $X_{\text{solute}} = 1 - X_{\text{solvent}} = 1 - 0.9 = 0.1$

   The molality $m$ of the solution is given by:
   $m = \frac{n_{\text{solute}}}{W_{\text{solvent}}\text{ (in kg)}} = \frac{X_{\text{solute}}}{X_{\text{solvent}} \times M_{\text{solvent}}}$

   Given that the solvent is water, the molar mass of water is $M_{\text{solvent}} = 18\text{ g mol}^{-1} = 18 \times 10^{-3}\text{ kg mol}^{-1}$.

   Substituting the values:
   $m = \frac{0.1}{0.9 \times 18 \times 10^{-3}} = \frac{1000}{162} \approx 6.173\text{ mol kg}^{-1}$

2. Calculate the Elevation in Boiling Point ($\Delta T_b$):
   The molal boiling point elevation constant for water is $K_b = 0.52\text{ K kg mol}^{-1}$.
   Assuming the solute is a non-electrolyte ($i = 1$):
   $\Delta T_b = K_b \times m$
   $\Delta T_b = 0.52\text{ K kg mol}^{-1} \times 6.173\text{ mol kg}^{-1} \approx 3.21\text{ K}$

3. Determine the Boiling Point of the Solution ($T_b$):
   The normal boiling point of pure water is $T_b^\circ = 373.0\text{ K}$ (or $373.15\text{ K}$).
   $T_b = T_b^\circ + \Delta T_b = 373.0\text{ K} + 3.21\text{ K} \approx 376.2\text{ K}$

Therefore, the boiling point of the solution is $376.2\text{ K}$, which corresponds to option (B).