Question

The term proof is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is 95% (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92-proof ethanol–water solution? Assume the density of ethanol $\left({\mathbf{C}}_2{\mathbf{H}}_5\mathbf{OH}\right)$ is 0.79 g/cm3 and the density of water is 1.0 g/cm3.

Short answer

The molarity is $0.17mol$.

Step-by-step solution

Step 1:

Prove is defined as twice the percent by volume of pure ethanol in solution. Hence, the percentage of ethanol in solution is$\omega \left(C_2{H}_{5}OH\right)=\frac{92}{2}=46\%=0.46$.

The density of ethanol is $0.79gram/c{m}^3$.

Step 2:

To calculate morality use the formula $m=\frac{\text{No.ofsolute}}{\text{Litresofsolution}}$ .

$\begin{array}{l}=\frac{m\left(C_2{H}_{5}OH\right)\cdot M\left(C_2{H}_{5}OH\right)}{100mL}\\ =\frac{\omega \left(C_2{H}_{5}OH\right)\cdot \rho \left(C_2{H}_{5}OH\right)\cdot \left(C_2{H}_{5}OH\right)}{0.1L}\\ =\frac{0.46\cdot 0.79g/c{m}^3\cdot 46.068g/mol}{100mL}\\ =0.17mol\end{array}$