Calculate the density `("in gm L"^(-1))` of a 3.60 M sulphuric acid solution that is 29 % `H_(2)SO_(4)` by mass `("molar mass = 98 g mol"^(-1))`

Options

A. 1.45
B. 1.64
C. 1.88
D. 1.22

Answer

D

Explanation

`29% H_(2)SO_(4)` by mass means 29 g `H_(2)SO_(4)` are present in 100 g of the solution. If `"d g mL"^(-1)` is the density of the solution, volume `=(100)/(d)mL.`
Further, `"29 g "H_(2)SO_(4)=(29)/(98)" mole."`
Hence, molarity of the solution
`=(29//98)/(100//d)xx"1000 mol L"^(-1) ="3.60 mol L"^(-1)" (Given)"`
`"i.e., "(29)/(98)xx(1)/(100)xx1000=3.60`
`"or "d=(3.60)/(10)xx(98)/(29)="1.22 g mL"^(-1)`