Calculate the mass of sodium acetate \((\text{CH}_3\text{COONa})\) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is \(82.0245\, \text{g mol}^{–1}.\)

Question

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:

Answer 1

Given

Molarity of solution, M = 0.375 mol·L⁻¹
Volume of solution, V = 500 mL = 0.500 L
Molar mass of sodium acetate (CH₃COONa) = 82.0245 g·mol⁻¹

Step 1: Calculate moles of sodium acetate

Moles of solute (n) = M × V = 0.375 × 0.500 = 0.1875 mol

Step 2: Calculate mass of sodium acetate

Mass = n × molar mass = 0.1875 × 82.0245 g = 15.38 g (approximately)

Required mass of sodium acetate ≈ 15.4 g.

Calculate the mass of sodium acetate \((\text{CH}_3\text{COONa})\) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is \(82.0245\, \text{g mol}^{–1}.\)

Solution and Explanation

Given

Step 1: Calculate moles of sodium acetate

Step 2: Calculate mass of sodium acetate

Top Questions on Mole concept and Molar Masses

Questions Asked in CBSE Class XI exam