Question

The elevation in boling point of an aqueous solutions of NaCl is 0.01 .∘C . If its van't Hoff factor is 1.92 ,the molality of NaCl solution is

( Kb for water = 0 . 52 k kg mol⁻¹)

A. 0.01m

B. 0.001m

C. 0.005m

D. 0.02m

Answer 1

Using the boiling point elevation equation and given values, the molality of the NaCl solution with a boiling point elevation of 0.01°C and van't Hoff factor of 1.92 is calculated to be approximately 0.01 m.

Step-by-step explanation:

To find the molality of the NaCl solution given the elevation in boiling point and the van't Hoff factor, we can use the formula for boiling point elevation, which is ΔT = i × Kb × m, where ΔT is the boiling point elevation, i is the van't Hoff factor, Kb is the molal boiling point elevation constant for water, and m is the molality of the solution. For water, Kb is known to be 0.51°C/m. We were given that the elevation in boiling point, ΔT, is 0.01°C and the van't Hoff factor, i, is 1.92. Plugging in these values into the equation, we get:

0.01°C = 1.92 × 0.51°C/m × m

Solving for molality m, we divide both sides by (1.92 × 0.51°C/m), which gives m = 0.01°C / (1.92 × 0.51°C/m).

Therefore, the molality of the NaCl solution is approximately 0.01°C / 0.9792°C/m, which is about 0.0102 m, or 0.01 m when rounded to two decimal places.

Hence, the correct answer is B, 0.01 m.