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 -MM1)×100 M = Molarity M1 = Molecular mass d = density = (2.05/(1000×1.02)-(2.05×60))×1000 = 2.28 mol kg-1

Q. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. 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) $= \frac{M}{1000 d - M M _{1}} \times 100$
M = Molarity
$M_{1}$ = Molecular mass
d = density
$= \frac{2.05}{( 1000 \times 1.02 ) - ( 2.05 \times 60 )} \times 1000$
$= 2.28 m o l k g^{- 1}$

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

1.14molkg−1

3.28molkg−1

2.28molkg−1

0.44molkg−1

molality (m) =1000d−MM1​M​×100 M = Molarity M1​ = Molecular mass d = density =(1000×1.02)−(2.05×60)2.05​×1000 =2.28molkg−1

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

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

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

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

molality (m) $= \frac{M}{1000 d - M M _{1}} \times 100$
M = Molarity
$M_{1}$ = Molecular mass
d = density
$= \frac{2.05}{( 1000 \times 1.02 ) - ( 2.05 \times 60 )} \times 1000$
$= 2.28 m o l k g^{- 1}$