Molarity from Mass Percentage and Density JEE Chemistry

 

❓ Question

The molarity of a 70% (m/m) aqueous solution of a monobasic acid (X) is:

Given:

• Density of solution = 1.25 g mL⁻¹
• Molar mass of acid = 70 g mol⁻¹

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✍️ Solution

Assume 100 g of solution.

→ Mass of acid (solute) = 70 g

→ Mass of water (solvent) = 30 g

Given,

$$\text{Density} = 1.25\ \text{g mL}^{-1}$$

Volume of solution,

$$V=\frac{\text{Mass}}{\text{Density}} =\frac{100}{1.25} =80\ \text{mL} =0.080\ \text{L}$$

Molar mass of acid,

$$M = 70\ \text{g mol}^{-1}$$

Number of moles of acid,

$$n=\frac{\text{Mass}}{\text{Molar mass}} =\frac{70}{70} =1\ \text{mol}$$

Now,

$$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$$
$$=\frac{1}{0.080} =12.5\ \text{M}$$

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✅ Final Answer

$$\boxed{12.5\ \text{M}}$$