JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Topic Test - Solutions

Find the molality of \[{{H}_{2}}S{{O}_{4}}\] solution whose specific gravity is \[1.98\,g\,m{{l}^{-}}\] and 95% of volume is \[{{H}_{2}}S{{O}_{4}}\].

A) 7.412

B) 8.412

C) 9.412

D) 10.412

Correct Answer: C

Solution :

[c] \[{{H}_{2}}S{{O}_{4}}\] is 95% by volume, so

wt. of \[{{H}_{2}}S{{O}_{4}}=95\,g\]

Vol. of solution = 100 ml

\[\therefore \] Moles of \[{{H}_{2}}S{{O}_{4}}=\frac{95}{98}.\]

And weight of solution \[=100\times 1.98=198\text{ }g\]

Weight of water \[=198-95=103\text{ }g\]

Molality \[=\frac{95\times 1000}{98\times 103}=9.412\]

Hence, molality of \[{{H}_{2}}S{{O}_{4}}\] solution is 9.412.