Question

density of 2.03 M aqueous solution of acetic acid is 1.017 g m/L molecular mass of acetic acid is 60. calculate the molality of solution

Answer 1

Answer:

Given,

Density=1.017gmL

−1

Molecular Mass=60

Molarity=2.03M

→Molarity=2.03M means 2.03 moles in 1L of solution.

n=

Molar Mass

Given Mass

2.03=

60

Given Mass

2.03×60=121.8g

121.8g of acetic acid in water.

→Density=1.017gmL

−1

1mLofsolution⟶1.017ofsolution

1000mLofsolution⟶1017gofsolution

Mass of solution = Mass of solvent + Mass of solute

1017g = Mass of solvent + 121.8g

1017−121.8g = Mass of solvent

895.2g = Mass of solvent

Molality=

Mass of solvent(kg)

no. of moles

=

895.2×10

−3

kg

2.03

=2.267m