Question

The density of 1.25M Na2SO4 solution of 1.25 g/ml calculate 1)mole fraction 2 )mass percent and molality of Na2SO4

Answer 1

Answer:

mole fraction = 0.02

% by mass = 14.2 %

molality = 1.17

Explanation:

Given: Density of Na2SO4 = 1.25 g/ml

Molarity of solution = 1.25 M (1.25 moles in 1L)

Moles of Na2SO4 = n = 1.25

Moles of water =

Solution:

density = mass / Volume

Weight of solution = Volume of solution X Density

= 1000ml X 1.25

= 1250 g

Moles = weight of solute / molar mass

Weight of solute = moles X molar mass

= 1.25 X 142

= 177.5

Weight of solution = wgt of solute + wgt of solvent

Weight of solvent = wgt of solution—wgt of solute

= 1250 — 177.5

= 1072.5 g

1. Mole fraction X(Solute) = n/(n+N)

moles of solvent = weight / Molar mass

= 1072.5/18

= 59.58

Mole fraction of solute X = 1.25/(1.25+59.58)

= 1.25/60.83

= 0.02

2. Mass % = (mass of solute/mass of solution)X100

= (177.5/1250)X100

= 14.2 %

3. Molality=(moles ofsolute/mass of solvent)X1000

= (1.25/1072.5)X1000

= 1.17 m