Question

What is mole fraction of 56% w/v KOH soln. having density 1 g/cm3

Answer 1

Hey Dear,

◆ Answer -

X2 = 0.29

◆ Explaination -

# Given -

w/v % = 56 %

d = 1 g/cm^3

# Solution -

56 % w/v means 56 g of KOH in 100 cm^3 solution.

As density is given -

100 cm^3 = 100 × 1 = 100 g

Now we have 56 g KOH in 100-56 = 44 g water.

No of moles of H2O in 100 g solution -

n2 = 44 / 18

n2 = 2.444 mol

No of moles of KOH in 100 g solution -

n2 = 56 / 56

n2 = 1 mol

Mole fraction of KOH -

X2 = n2 / (n1 + n2)

X2 = 1 / (2.444 + 1)

X2 = 0.29

Therefore, mole fraction of solute in KOH soln is 0.29.

Hope this helps you.