Question

Molarity of $H_{2}S{O}_4$ (density 1.8 g / mL) is 18 M. The molality of this $H_{2}S{O}_4$ solution is:

• A. $36$
• B. $200$
• C. $500$
• D. $18$

Answer 1

The correct option is C $500$

$Given$ : Molarity of $H_{2}S{O}_4$ = $18M$

As, Molarity = $\frac{no.ofmolesofsolute}{volumeofsolution}$

i.e., $18moles$ of sulphuric acid is present per litre of solution.

$Density=1.8g/mL$ {given}

Thus , mass of one litre of this solution ,

Density = $\frac{mass}{volume}$

mass = $density\times volume$

mass = $1.8\times 1000$

mass = $1800g$

Thus , mass of $1litre$ $18M$ $H_{2}S{O}_4$ solution is $1800g$ or $1.8kg$.

Now, mass of solution = mass of solute + mass of solvent

mass of solution = no. of moles of solute X molar mass of solute + mass of solvent

$1800$ = $18\times 98$ + mass of solvent

mass of solvent = $(1800-1764)g$

mass of solvent = $36g$

Now , molality of $H_{2}S{O}_4$ solution is given by,

molality = $\frac{numberofmolesofsolute}{massofsolvent\left(inkg\right)}$

molality = $\frac{18\times 100}{36}$

molality = $500m$

Hence, the molality of $H_{2}S{O}_4$ solution is $500m$.