Question

At 298 K, the density of an aqueous solution containing 82 g of acetic acid per dm\(^3\) is 1.01 kg dm\(^{-3}\). If the molarity of the solution is \(x\) M, the molality \(m\) of the same solution is (molar mass of acetic acid = 60 g mol\(^{-1}\)):

• A. \( (1.856 \times x) \) m
• B. \( (0.999 \times x) \) m
• C. \( (0.928 \times x) \) m
• D. \( (1.077 \times x) \) m

Hint:

The relationship between molarity and molality depends on the density of the solution. Be sure to use the appropriate conversion factor when calculating these values.

Answer 1

The Correct Option is D

Step 1: Calculate the molarity \( x \). We are given that 82 g of acetic acid is dissolved in 1 dm\(^3\) of solution. Using the molar mass of acetic acid \( 60 \, {g/mol} \), the molarity is: \[ x = \frac{82 \, {g}}{60 \, {g/mol}} = 1.3667 \, {M}. \] Step 2: Use the formula for molality in terms of molarity. We know the molarity \( x \) is related to molality \( m \) by the equation: \[ m = \frac{x \times {density of solution (kg/dm}^3)}{1000}. \] Substitute the given values into this equation: \[ m = \frac{1.3667 \times 1.01}{1000} = 0.00138 \, {mol/kg}. \] Thus, \( m = 1.077 \times x \).