Molality (m) of 3 M aqueous solution of NaCl is:

Question

Molality (m) of 3 M aqueous solution of NaCl is:
(Given: Density of solution = 1.25 g ${mL}^{- 1}$ , Molar mass in g ${mol}^{- 1}$ : Na - 23, Cl - 35.5) [2024]

Smart Achievers · Accepted Answer

(2)

3M aqueous NaCl means:

3 mol NaCl in 1000 mL solution

$3 \times 58.5 g = 175.5 g NaCl in 1.25 {g mL}^{- 1} \times 1000 mL - 1250 g solution$

$3 mol NaCl in (1250 - 175.5) g = 1074.5 g = 1.074 kg solution$

$m = \frac{n_{solute}}{V_{solvent} (in kg)} = \frac{3}{1.0745} m = 2.79 m$

Solution

Q.
Molality (m) of 3 M aqueous solution of NaCl is:
(Given: Density of solution = 1.25 g ${mL}^{- 1}$ , Molar mass in g ${mol}^{- 1}$ : Na - 23, Cl - 35.5) [2024]

1 2.90 m

2 2.79 m

3 1.90 m

4 3.85 m

Ans.
(2)

3M aqueous NaCl means:

3 mol NaCl in 1000 mL solution

$3 \times 58.5 g = 175.5 g NaCl in 1.25 {g mL}^{- 1} \times 1000 mL - 1250 g solution$

$3 mol NaCl in (1250 - 175.5) g = 1074.5 g = 1.074 kg solution$

$m = \frac{n_{solute}}{V_{solvent} (in kg)} = \frac{3}{1.0745} m = 2.79 m$

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