Question

what is the molarity of a 750.00 ml solution which contains 164.5 g of sodium sulfate, na₂so₄? (mm = 142.04 g mol⁻¹)
○ 1.544 m
○ 0.001544 m
○ 0.2193 m
○ 1.2372 m

Explanation:

Step1: Calculate moles of $\ce{Na2SO4}$

Moles = $\frac{\text{mass}}{\text{molar mass}}$. Mass = 164.5 g, molar mass = 142.04 g/mol.
Moles = $\frac{164.5}{142.04} \approx 1.158$ mol.

Step2: Convert volume to liters

Volume = 750.00 mL = $\frac{750.00}{1000} = 0.75$ L.

Step3: Calculate molarity

Molarity (M) = $\frac{\text{moles of solute}}{\text{volume of solution (L)}}$.
M = $\frac{1.158}{0.75} \approx 1.544$ M.

Answer:

1.544 M