Question

A solution of methanol in water is 20 % by volume. If the solution and pure methanol have densities of \( 0.964\text{ kg L}^{-1} \) and \( 0.793\text{ kg L}^{-1} \), respectively, find the per cent of methanol by weight?

• A. 15.8
• B. 16.45
• C. 20
• D. 14.8
• E. 17.6

Hint:

To convert volume % to weight %, you must use the specific densities of both the solute and the entire solution to find their respective masses.

Answer 1

The Correct Option is B

Concept: The percent by weight is the mass of the solute divided by the total mass of the solution, multiplied by 100. Mass can be calculated using the density formula: $\text{mass} = \text{density} \times \text{volume}$.

Step 1: {Assume a standard volume for the solution.} Let the total volume of the solution be $V_{\text{sol}} = 1\text{ L}$. Given 20 % methanol by volume: $$\text{Volume of methanol } (V_{\text{meth}}) = 0.20 \times 1\text{ L} = 0.20\text{ L}$$

Step 2: {Calculate the mass of methanol and the mass of the solution.} $$\text{Mass of methanol } (m_{\text{meth}}) = \text{Density}_{\text{meth}} \times V_{\text{meth}}$$ $$m_{\text{meth}} = 0.793\text{ kg/L} \times 0.20\text{ L} = 0.1586\text{ kg}$$ $$\text{Mass of solution } (m_{\text{sol}}) = \text{Density}_{\text{sol}} \times V_{\text{sol}}$$ $$m_{\text{sol}} = 0.964\text{ kg/L} \times 1\text{ L} = 0.964\text{ kg}$$

Step 3: {Calculate the weight percentage.} $$\text{Weight \%} = \frac{m_{\text{meth}}}{m_{\text{sol}}} \times 100$$ $$\text{Weight \%} = \frac{0.1586}{0.964} \times 100$$ $$\text{Weight \%} \approx 16.452\%$$