Question

question 4 (2 points) the molality of a solution that contains 36.6 g of nacl in 200.0 g of water is ______ m. (molar mass of nacl = 58.44 g/mol.) a) 3.66 b) 3.13 c) 31.3 d) 36.6

Explanation:

Step1: Calculate moles of NaCl

Moles = $\frac{\text{Mass}}{\text{Molar Mass}}$ = $\frac{36.6\ \text{g}}{58.44\ \text{g/mol}}$ ≈ 0.626 mol

Step2: Convert water mass to kg

200.0 g = 0.2000 kg

Step3: Calculate molality

Molality (m) = $\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$ = $\frac{0.626\ \text{mol}}{0.2000\ \text{kg}}$ ≈ 3.13 m

Answer:

b) 3.13