Question

An aqueous solution of a dibasic acid (MW = 118) containing 35.4 g of the acid per liter of the solution, has density 1.00 g/cm\(^3\). The normality and molality values of the solution respectively will be:

• A. 0.6 N, 0.3 mol Kg\(^-1\)
• B. 0.3 N, 0.3 mol Kg\(^-1\)
• C. 0.3 N, 0.6 mol Kg\(^-1\)
• D. 0.6 N, 0.6 mol Kg\(^-1\)

Hint:

For dibasic acids, remember to divide the molecular weight by 2 to find the equivalent weight when calculating normality. Molality is calculated using the mass of the solvent, not the solution.

Answer 1

The Correct Option is A

Given that the molecular weight (MW) of the dibasic acid is 118, the mass of the acid is 35.4 g, and the density of the solution is 1.00 g/cm\(^3\), we can calculate the normality and molality.
Step 1: Calculating Normality (N) Normality is given by the formula: \[ N = \frac{\text{Gram equivalent weight}}{\text{Volume of solution in liters}} \] The equivalent weight of the acid is the molecular weight divided by 2, as it is dibasic. So, \[ \text{Equivalent weight} = \frac{118}{2} = 59 \, \text{g/equivalent} \] The normality is: \[ N = \frac{35.4}{59 \times 1} = 0.6 \, \text{N} \] Step 2: Calculating Molality (m) Molality is calculated using: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] First, we calculate the moles of solute: \[ \text{Moles of acid} = \frac{35.4}{118} = 0.3 \, \text{mol} \] Since the density of the solution is 1.00 g/cm\(^3\), 1 liter of solution weighs 1000 g. Therefore, the mass of the solvent is: \[ \text{Mass of solvent} = 1000 \, \text{g} - 35.4 \, \text{g} = 964.6 \, \text{g} = 0.9646 \, \text{kg} \] Now, the molality is: \[ m = \frac{0.3}{0.9646} = 0.3 \, \text{mol Kg}^{-1} \] Thus, the normality and molality of the solution are 0.6 N and 0.3 mol Kg\(^-1\), respectively.