Question

The molality of $1 \,L$ solution with $x \% H _{2} SO _{4}( w / v )$ is equal to $9 .$ The weight of solvent present in the solution is $910 g$. The value of $x$ is-

• A. $90$
• B. $80.3$
• C. $100$
• D. $95.5$

Answer 1

Answer: (B)

molality $=9$ which means,

9 moles of solute $\left( H _{2} SO _{4}\right)$ are present per $kg$ of solvent.

So, no. of moles of solute in $1000 g$ solvent $=9$

no. of moles of solute in $910 g$ solvent $=\frac{9}{1000} \times 910$

$=\frac{819}{100}$

$=8.19 $ moles

wt. of 8.19 moles of $H _{2} SO _{4}=98 \times 8.19=802.62 g$

$\%$ age $( w / v )$ of $H _{2} SO _{4}=\frac{802.62 g }{1000 ml } \times 100=80.26 \%$