In what ratio must distilled water and alcohol be mixed to…

In what ratio must distilled water and alcohol be mixed to obtain a liquid of density $\rho = 800\ \mathrm{kg/m^3}$? When water and alcohol are mixed, the total volume decreases by 3% compared with the sum of the volumes of the components. (Densities: water $\rho_w = 1000$, alcohol $\rho_a = 789\ \mathrm{kg/m^3}$.)

Explanation of Method

Mixing water and alcohol to a given density.

Step 1: Mass and volume of the mixture

\[ m = \rho_w V_w + \rho_a V_a, \]

\[ V = 0.97\,(V_w + V_a). \]

Step 2: Condition for the density

\[ 800 \cdot 0.97\,(V_w + V_a) = 1000\,V_w + 789\,V_a, \]

\[ 776\,V_w + 776\,V_a = 1000\,V_w + 789\,V_a. \]

\[ -224\,V_w = 13\,V_a \;\Longrightarrow\; \frac{V_w}{V_a} = -\frac{13}{224}. \]

The ratio comes out negative, so no real mixture works. The reason is simple: after a 3% shrinkage even alcohol alone would have $789/0.97 \approx 813\ \mathrm{kg/m^3}$, and adding water only makes the mixture denser.

\[\begin{gathered}\text{With a 3\% shrinkage,} \\ \rho = 800\ \mathrm{kg/m^3} \text{ cannot be reached} \end{gathered}\]

$0$ $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $800$ $850$ $900$ $\rho_{\text{mix}}\ (\mathrm{kg/m^3})$ $\text{water fraction } \frac{V_w}{V_w + V_a}$ $\text{target } 800$ $11 : 200$ $\text{with shrinkage: always} > 813$

no volume change
volume shrinks by 3%

Mixture density against the share of water. Without shrinkage the line crosses 800 kg/m³ at about 5% water; with a 3% shrinkage it starts at 813 kg/m³ and never gets down to 800.

Step 3: Without the shrinkage

If the volumes simply add up: $800\,(V_w + V_a) = 1000\,V_w + 789\,V_a \;\Longrightarrow\; 11\,V_a = 200\,V_w$.

\[V_w : V_a = 11 : 200 \approx 1 : 18\]

General formula

With a shrinkage factor $k$ (here $k = 0.97$): $\frac{V_w}{V_a} = \frac{k\rho - \rho_a}{\rho_w - k\rho}$. which is positive only when $k\rho \gt \rho_a$, i.e. here for $\rho \gt 813\ \mathrm{kg/m^3}$. For example, $\rho = 900\ \mathrm{kg/m^3}$ would need $V_w : V_a = 84 : 127$.

Result

\[\boxed{\,\begin{gathered}\text{With a 3\% shrinkage,} \\ \rho = 800\ \mathrm{kg/m^3} \text{ cannot be reached} \end{gathered}\,}\]

\[\boxed{\,V_w : V_a = 11 : 200 \approx 1 : 18\,}\]

Remark

About the original solution. Its algebra is right up to $-224\,V_w = 13\,V_a$, but it then dropped the minus sign and gave $13 : 224$. That ratio does not satisfy the equation, and a negative volume ratio means the requested mixture is impossible with the given data. (In reality, the volume shrinkage for such a small amount of water would also be much smaller than 3%.) Most likely the problem intends either no shrinkage (answer 11 : 200) or a higher target density.