Chemistry: An aqueous solution that is 50.0 percent sulfuric acid (h2so4) by mass has a density of 1.143 g/ml. determine the molality of the solution.

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An aqueous solution that is 50.0 percent sulfuric

ana9340

12.03.2022

Molality is defined as the ratio of number of moles of solute to the mass of solvent in kilograms.

The formula of molality is:

$Molality = \frac{number of moles of solute}{mass of solvent in kg}$ - (1)

The formula of density is:

$Density = \frac{mass}{volume}$ - (2)

Consider, Volume = 1 L = 1000 mL

Rearranging the formula (2):

$mass of solution = density \times volume$

Substituting the values:

$mass of solution = 1.143 g/mL \times 1000 mL = 1143 g$

Since, the aqueous solution is 50% by mass that is:

$50 = \frac{mass of H_2SO_4}{mass of solution}\times 100$

Rearranging the equation:

$mass of H_2SO_4 = \frac{50}{100}\times mass of solution$

$mass of H_2SO_4 = \frac{50}{100}\times 1143 = 571.5 g$

Now, for determining the number of moles of H_2SO_4 :

$n_{H_2SO_4 } = \frac{mass of H_2SO_4 }{Molar mass of H_2SO_4 }$

$n_{H_2SO_4 } = \frac{571.5 g}{98.08 g/mol} = 5.83 moles$

mass of solvent = mass of solution - mass of solute

mass of solvent = 1143 g - 571.5 g = 571.5 g = 0.5715 kg

Substituting the values in formula (1):

$molality = \frac{5.83 moles}{0.5715 kg} = 10.20 m$

Hence, the molality of the solution is 10.20 m .

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