Question

Determine the molality of an aqueous solution that is 11.5 percent urea by mass.

__ m

Determine the percent sulfuric acid by mass of a 1.67 m aqueous solution of H₂SO₄.

__ %

Answer 1

1)

Urea is: CH4N2O

Let mass of solution be 1 Kg = 1000 g

mass of CH4N2O = 11.5 % of mass of solution

= 11.5*1000.0/100

= 115.0 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 115.0 g

mass of solvent = 885.0 g

mass of solvent = 0.885 Kg

Molar mass of CH4N2O,

MM = 1*MM(C) + 4*MM(H) + 2*MM(N) + 1*MM(O)

= 1*12.01 + 4*1.008 + 2*14.01 + 1*16.0

= 60.062 g/mol

mass(CH4N2O)= 115.0 g

use:

number of mol of CH4N2O,

n = mass of CH4N2O/molar mass of CH4N2O

=(1.15*10^2 g)/(60.06 g/mol)

= 1.915 mol

m(solvent)= 0.885 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.915 mol)/(0.885 Kg)

= 2.163 molal

Answer: 2.16 molal

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