Molality (m) of 3M 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)
- 2.90 m
- 3.85 m
- 1.90 m
- 2.79 m
Answer
(4) 2.79 m
Solution
Take 1 L solution: mass = 1000 1.25 = 1250 g
NaCl = 3 mol = 3 58.5 = 175.5 g
⇒ water = 1250 - 175.5 = 1074.5 g = 1.0745 kg
∴ m = 31.0745 = 2.79 m
Answer: (4)