At room temperature, the density of water is $1.0\text{g/mL}$ and the density of ethanol is $0.789\text{g/mL}$. What volume of ethanol contains the same number of molecules as are present in $175\text{mL}$ of water?

The correct option is D $566.82\text{mL}$
Let the volume of ethanol containing the same number of molecules as are present in $175\text{mL}$ of $H_{2}O$ be $\text{v mL}$.
Thus,
$\text{moles of}{C}_2{H}_{5}OH\text{in v mL}=\text{moles of}{H}_{2}O\text{in 175 mL}$
Therefore,

$\frac{\text{weight of}{C}_2{H}_{5}OH}{\text{molar mass of}{C}_2{H}_{5}OH}=\frac{\text{weight of}{H}_{2}O}{\text{molar mass of}{H}_{2}O}$

$\Rightarrow \frac{0.789\times \text{v}}{46}=\frac{1.0\times 175}{18}$

$\Rightarrow \text{v}=566.82\text{mL}$