Question

1 kg of 2m Urea solution is mixed with 2 kg of 4m Urea solution the molality of the resulting solution is

Answer 1

Answer:

The molality of the resulting solution is 4.1672 mol/kg.

Explanation:

Moles of urea in 1 kg 2 m solution.

$2m=\frac{n}{1 kg}$

Moles of urea in 2 m solution ,n = 2 mol

Mass of 2 moles of urea = 2 × 60.06 g/mol=120.12 g

Mass of solvent = 1kg -0.12012 kg =0.87988 kg

Moles of urea in 2kg 4 m solution.

$4m=\frac{n'}{2 kg}$

Moles of urea in 4 m solution ,n' = 8 mol

Mass of 8 moles of urea = 8 × 60.06 g/mol= 480.24 g

Mass of solvent = 2 kg - 0.48024 kg = 1.51976 kg

After mixing of two solution the new molality of solution will be:

Total moles of urea = n+ n'= 2 mol + 8 mol = 10 mol

Mass of solvent after mixing =0.87988 kg + 1.51976 kg =2.39964 kg

$m=\frac{10 mol}{2.39964 kg}=4.1672 kg$

The molality of the resulting solution is 4.1672 mol/kg.

Answer 2

Answer:

m={(Mass1×molality1)+(mass2×molality2)}÷mass1+mass2

={(1×2)+(2×4)}÷1+2

={2+8}÷3

=10÷3

=3.33

Hope you understood

Thank you