Question

A 68.8 g H2SO4(MM = 98 g/mol) is added to H2O to make a 500 mL solution. If the density of the solution is 1.2 g/mL, calculate the molality.

Answer 1

Answer:

Molar mass of H

2

SO

4

=98 g

Volume of solution =500ml=0.5l

Molarity =

Volume of solution

Moles of solute

=

98×0.5

98

=2 M

Answer 2

The molality of the solution is 1.32 molal.

Given,

Mass of H2SO4=68.8 g

Volume of the solution=500 mL

Density of the solution=1.2 g/mL.

To find,

the molality of the solution.

Solution:

• Molality of a substance in as solution is defined as the moles of that substance in 1 Kg of solvent.
• Molality of a substance is temperature independent.
• Its unit is moles/Kg or molal.
• Molality=(Moles/Mass of solvent(Kg)).

The molar mass of H2SO4 is 98 g.

The mass of the solute(H2SO4) is 68.8 g.

Moles of H2SO4 in the solution will be:

$Moles=\frac{Mass}{Molar mass}\\Moles=\frac{68.8}{98}\\Moles=0.702.$

Let the mass of the solvent be w.

Mass of the solution=Mass of solute+Mass of solvent

Mass of the solution=(68.8+w) g.

The density of the solution is 1.2 g/mL.

The mass of the solvent will be:

$Density=\frac{Mass}{Volume}\\1.2=\frac{68.8+w}{500}\\1.2X500=68.8+w\\600=68.8+w\\w=(600-68.8)g\\w=531.2 g\\531.2 g=\frac{531.2}{1000} \\531.2 g=0.5312 Kg\\531.2g=0.531 Kg.$

The mass of the solvent is 0.53 Kg.

Molality=(Moles/Mass of solvent(Kg))

$Molality=\frac{0.702}{0.531} \\Molality=1.32 moles/Kg.$

Hence, the molality is 1.32 molal.

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