1.02 g of the following substances are dissolved in 100 mL of water.…

The correct answer is Option (1) → (C), (B), (A), (D)

Step 1: Recall the formula for molarity

$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$

Volume = 100 mL = 0.1 L

Moles = $\frac{\text{mass}}{\text{molar mass}}$

Step 2: Calculate molar mass and moles

NaCl → Molar mass = 58.44 g/mol

$n = \frac{1.02}{58.44} \approx 0.01745 \, \text{mol}$

$M=\frac{0.01745}{0.1} \approx 0.1745 \, \text{M}$

KCl → Molar mass = 74.55 g/mol

$n = \frac{1.02}{74.55} \approx 0.01368 \, \text{mol}$

$M=\frac{0.01368}{0.1} \approx 0.1368 \, \text{M}$

MgSO₄ → Molar mass = 120.37 g/mol

$n = \frac{1.02}{120.37} \approx 0.00848 \, \text{mol}$

$M=\frac{0.00848}{0.1} \approx 0.0848 \, \text{M}$

NaOH → Molar mass = 40 g/mol

$n = \frac{1.02}{40} = 0.0255 \, \text{mol}$

$M=\frac{0.0255}{0.1} = 0.255 \, \text{M}$

Step 3: Arrange in increasing order

MgSO₄ (0.0848) < KCl (0.1368) < NaCl (0.1745) < NaOH (0.255)