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MaRussiya [10]
3 years ago
6

A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

ulate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.
Chemistry
1 answer:
rodikova [14]3 years ago
6 0

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = 1 g/cm^3=1 g/mL

1 mL= 1 cm^3

m=d\times v=1.0 g/mL\times 100.0 = 100.0 g

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol

Moles of water  = n_2=\frac{100.0 g}{18g/mol}=5.556 mol

Mole fraction of phosphoric acid =\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}

\chi_1=0.01803

Mole fraction of water =\chi_2

\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}

\chi_2=0.9820

[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}

Moles of  phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

=\frac{0.1020 mol}{0.113 L}=0.903 M

[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}

Moles of  phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg

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