The boiling point of a $4 \%$ aqueous solution of a non-volatile solute $\mathrm{P}$ is equal to the boiling point of $\mathrm{X} \%$ solution of another non-volatile solute $\mathrm{Q}$. The relation between their Molar masses is $\mathrm{M}_{\mathrm{Q}}=4 \mathrm{~Mp}$. What is $\mathrm{X}$ ?

The elevation in boiling point is given by $\Delta T_b = K_b \times m$, where $m$ is the molality of the solution. Since the boiling points of the two solutions are equal, their elevations in boiling point are equal, which implies their molalities must be equal, assuming the solvent is the same and the solutes are non-volatile and non-electrolytic.

Let $w_P$ and $w_Q$ be the masses of solutes P and Q in $100 \text{ g}$ of solution. For a $4 \%$ solution of P, $w_P = 4 \text{ g}$ and the mass of solvent is $100 - 4 = 96 \text{ g}$. The molality $m_P$ is given by $m_P = \dfrac{w_P / M_P}{W_{solvent, P} / 1000} = \dfrac{4 / M_P}{96 / 1000} = \dfrac{4000}{96 M_P} = \dfrac{125}{3 M_P}$.

For an $X \%$ solution of Q, $w_Q = X \text{ g}$ and the mass of solvent is $100 - X \text{ g}$. The molality $m_Q$ is given by $m_Q = \dfrac{X / M_Q}{(100 - X) / 1000} = \dfrac{1000 X}{(100 - X) M_Q}$.

Equating $m_P$ and $m_Q$: $\dfrac{125}{3 M_P} = \dfrac{1000 X}{(100 - X) M_Q}$.

Given $M_Q = 4 M_P$, substitute this into the equation: $\dfrac{125}{3 M_P} = \dfrac{1000 X}{(100 - X) (4 M_P)}$.

Simplifying: $\dfrac{125}{3} = \dfrac{250 X}{100 - X} \Rightarrow \dfrac{1}{3} = \dfrac{2 X}{100 - X} \Rightarrow 100 - X = 6 X \Rightarrow 7 X = 100 \Rightarrow X = \dfrac{100}{7} \approx 14.2857$.

Rounding to two decimal places, $X = 14.29$.

Answer: 14.29