At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mmHg of n-heptane in its pure state?

Let us denote n-hexane as component A and n-heptane as component B. At 300 K the solution is assumed to behave ideally, so Raoult’s law applies.

Raoult’s law (for an ideal binary solution) states

$$P_{\text{total}} \;=\; x_A P_A^{*} \;+\; x_B P_B^{*}$$

where $$x_A=\frac{\text{moles of A}}{\text{total moles}}, \qquad x_B=\frac{\text{moles of B}}{\text{total moles}},$$ and $$P_A^{*},\;P_B^{*}$$ are the vapour pressures of the pure liquids A and B respectively.

First mixture: 1 mol of A and 3 mol of B (total = 4 mol).

$$x_A^{(1)}=\frac{1}{4}=0.25, \qquad x_B^{(1)}=\frac{3}{4}=0.75$$

The given total vapour pressure is 550 mmHg, so

$$0.25\,P_A^{*} \;+\; 0.75\,P_B^{*}=550\;.$$

Second mixture: one more mole of B is added, so we now have 1 mol of A and 4 mol of B (total = 5 mol).

$$x_A^{(2)}=\frac{1}{5}=0.20, \qquad x_B^{(2)}=\frac{4}{5}=0.80$$

The new total vapour pressure is 560 mmHg, hence

$$0.20\,P_A^{*} \;+\; 0.80\,P_B^{*}=560\;.$$

We now have two simultaneous linear equations:

$$\begin{aligned} (1)\;&\;0.25\,P_A^{*}+0.75\,P_B^{*}&=550,\\[4pt] (2)\;&\;0.20\,P_A^{*}+0.80\,P_B^{*}&=560. \end{aligned}$$

To remove decimals, rewrite them as fractions:

$$\begin{aligned} \frac14 P_A^{*}+\frac34 P_B^{*}&=550,\\[4pt] \frac15 P_A^{*}+\frac45 P_B^{*}&=560. \end{aligned}$$

Multiply each equation by 20 to clear denominators:

$$\begin{aligned} 5P_A^{*}+15P_B^{*}&=11000,\\[4pt] 4P_A^{*}+16P_B^{*}&=11200. \end{aligned}$$

Subtract the first transformed equation from the second:

$$\bigl(4P_A^{*}+16P_B^{*}\bigr)-\bigl(5P_A^{*}+15P_B^{*}\bigr)=11200-11000,$$

$$-P_A^{*}+P_B^{*}=200,$$

so

$$P_B^{*}=P_A^{*}+200\;.$$

Substitute this relation back into the first transformed equation:

$$5P_A^{*}+15\bigl(P_A^{*}+200\bigr)=11000,$$

$$5P_A^{*}+15P_A^{*}+3000=11000,$$

$$20P_A^{*}=11000-3000=8000,$$

$$P_A^{*}=\frac{8000}{20}=400\;\text{mmHg}.$$

Using $$P_B^{*}=P_A^{*}+200$$ gives

$$P_B^{*}=400+200=600\;\text{mmHg}.$$

Thus, the vapour pressure of pure n-heptane (component B) at 300 K is

$$600\;\text{mmHg}.$$

So, the answer is $$600$$.