The density (in g `mL^(-1)` ) of a 3.60 M sulphuric acid solution that is 29% `H_2SO_4` (molar mass `= 98 g mol^(-1)` ) by mass, will be

Options

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

Answer

D

Explanation

3.6 M solution means 3.6 moles of `H_2SO_4` is present in 1000 mL of solution.
`implies " Mass of 3.6 moles of " H_2SO_4 = 3.6 xx 98 g = 352.8 g`
`implies " Mass of " H_2SO_4 " in 1000 mL of solution " = 352.8 g`
Given, 29 g of `H_2SO_4` is present in 100 g of solution `(29% H_2SO_4` by mass)
`= 352.8 g of H_2SO_4 " is present in " 100/29 xx 352.8=1216 g` of solution
Now density `=("Mass")/("Volume") = (1216)/1000 = 1.216 g//mL = 1.22 g//mL`