Density of a 2.05 M solution of acetic acid in water is 1.02 g / mL. The molality of the solution is

Question

Density of a 2.05 M solution of acetic acid in water is 1.02 g / mL. The molality of the solution is (A) 1.14 mol kg -1 (B) 3.28 mol kg -1 (C) 2.28 mol kg -1 (D) 0.44 mol kg -1

Tardigrade Admin · Accepted Answer

Molality , m=(M/1000 d-M M2) × 1000 where, M= molarity, d= density, M2= molecular mass m=(2.05/1000 × 1.02-2.05 × 60)=(2.05/897) =2.28 × 10-3 mol g -1=2.28 mol kg -1

Q. Density of a $2.05 M$ solution of acetic acid in water is $1.02 g / m L$ . The molality of the solution is

$1.14 m o l k g^{- 1}$

$3.28 m o l k g^{- 1}$

$2.28 m o l k g^{- 1}$

$0.44 m o l k g^{- 1}$

Molality $, m = \frac{M}{1000 d - M M _{2}} \times 1000$

where, $M =$ molarity, $d =$ density, $M_{2} =$ molecular mass

$m = \frac{2.05}{1000 \times 1.02 - 2.05 \times 60} = \frac{2.05}{897}$

$= 2.28 \times 1 0^{- 3} m o l g^{- 1} = 2.28 m o l k g^{- 1}$

Q. Density of a 2.05M solution of acetic acid in water is 1.02g/mL. The molality of the solution is

1.14molkg−1

3.28molkg−1

2.28molkg−1

0.44molkg−1