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Oksi-84 [34.3K]

4 years ago

15

A sulfuric acid solution containing 571.6 g of H2SO4 per liter of aqueous solution has a density of 1.329 g/cm^3.Calculate:a. Ma

ss percentageb. Mole fractionc. Molalityd. molarity of H2SO4 in this solution.

1 answer:

12345 [234]4 years ago

8 0

Given :

Mass of $H_2SO_4$ is 571.6 g per liter .

Density of solution , $\rho=1.329\ g/cm^3$ .

To Find :

a. Mass percentage

b. Mole fraction

c. Molality

d. molarity of H2SO4 in this solution.

Solution :

Molar mass of $H_2SO_4$ , m = 1329 g/mol .

a ) Mass of $H_2SO_4$ contain in 1 liter is 1329 g .

$mass \ \%=\dfrac{571.6}{1329}\times 100=43.01 \%$

b ) Moles of $H_2SO_4$ = $\dfrac{571.6\ g}{98\g/mol}=5.83\ mol$ .

Moles of $H_2O$ = $\dfrac{1329-571.6\ g}{18\ g/mol}=42.08\ mol$ .

Mole fraction $=\dfrac{5.83}{5.83+42.08}=0.12$ .

c ) Molarity of $H_2SO_4$ $=\dfrac{5.83\ mol}{1 \ L}=5.83\ M$ .

Hence , this is the required solution .

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Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).

In this question, solute is dimethyl ether while the solvent is methanol.

Mole (n) = mass (M) ÷ molar mass (MM)

Mole of solute (dimethyl ether) = 148.5 ÷ 46.07

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Mole of solvent (methanol) = 90 ÷ 32.04

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