What volume of 95% `H_(2) SO_(4)` by weight `(d = 1.85 g mL^(-1))` and what mass of water must be taken to prepare `100 mL` of 15% solution of `H_(2) SO_(4)` `(d = 1.10 g mL^(-1))`

Options

A. 5 cc
B. 7.5 cc
C. 9.4 cc
D. 12.4 cc.

Answer

C

Explanation

Molarity of 95% `H_(2)SO_(4)`
`M=(W_(2) xx 1000)/(M_(2) xx V(cm)^(3))`
Molarity of 15%
`H_(2)SO_(4)=(15 xx 1000)/(98(100//1.85))=1.68M`
`underset((95%H_(2)SO_(4)))(M_(1)V_(1))= underset((15%H_(2)SO_(4)))(M_(2)V_(2))`