Conc. `H SO_4` has density of `1.9 g mL^(-1)` and is 99% by weight. Calculate the molarity of `H_2SO_4 " (Mol wt of " H_2SO_4 = 98 g mol^(-1)` ).

Options

A. 2.4 M
B. 24.2 M
C. 19.1 M
D. 9.8 M

Answer

C

Explanation

99% by weight `H_2 SO_4`, means 99 g of `H_2SO_4` is present in 100 g solution
`V_("solution") = M/d = 100/1.9 mL`
Molarity `= ("Moles of " H_2SO_4)/("Volume of solution (mL)")xx1000`
Moles of `H_2SO_4 =99/98`
Molarity `= (99//98)/(100//1.9)xx1000 = (99xx1.9)/(98xx100)xx1000=19.1M`