Question

Molarity of H2SO4 is 0.8 and its density is 1.06 g/cm3. What will be its concentration in terms of molality and mole fraction ?
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Answer 1

$\fbox{\texttt{\green{Answer:}}}$

Molality = $\bold{0.815\:mol\:kg^{-1}}$

Mole fraction = 0.014

$\fbox{\texttt{\pink{Given:}}}$

Molarity of $\bold{H_2SO_4}$ = 0.8. This means that 0.8 mole $\bold{H_2SO_4}$ are present in 1 L of the solution.

Density of $\bold{H_2SO_4}$= 1.06 g/cm³

$\fbox{\texttt{\blue{To\:find:}}}$

Molality and mole fraction of the solute.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

We know 0.8 mole $\bold{H_2SO_4}$ are present in 1 L of the solution (given).

0.8 mole of $\bold{H_2SO_4}$ = 0.8 × 98 g = 78.4 g

1 L $\bold{H_2SO_4}$ solution = 1000 × 1.06 g = 1060 g

$\therefore$ Mass of solvent (water) = 1060 - 78.4 g = 981.6 g =

$\bold{\frac{981.6}{18}moles=54.53\:moles}$

$\hrulefill$

Thus, 981.6 g of water contain 0.8 mole of $\bold{H_2SO_4}$.

$\therefore$ Molality =

$\bold{\frac{0.8\:mol}{981.6\:g}\times{}1000\:g\:kg^{-1}=0.815\:mol\:kg^{-1}}$

Further, the solution contains 0.8 mole of solute in 54.53 moles of solvent.

$\therefore$ Mole fraction of solute =

$\bold{\frac{0.8}{0.8+54.53}=0.014}$