$20\text{g}$ of non-electrolyte, non-volatile solute $\left(C_x{H}_{2x}{O}_x\right)$ when dissolved in $100\text{g}$ water at ${100}^{\circ }\text{C}$, lowers the vapour pressure of solution by $\left(\frac{1}{100}\right)th$ of the vapour pressure of pure water. What is formula of the compound?

The correct option is B $C_{12}{H}_{24}{O}_{12}$
Given, weight of solvent $\left(W_1\right)=100\text{g}$ and weight of solute $\left(W_2\right)=20g$
$p_1^0$ is the vapour pressure of pure water and $p_1$ is the vapour pressure of the solution.
We know, $\frac{p_1^0-p_1}{p_1^0}=\frac{w_2\times m_1}{m_2\times w_1}$
$\Rightarrow \frac{100-99}{100}=\frac{20\times 18}{30x\times 100}\Rightarrow x=12$
So, the formula is $C_{12}{H}_{24}{O}_{12}$.