A solution contains 441.0 gof `H_2SO_4` in 1 L of solution. If the density of solution is `"1.25 g mL"^(-1),` the molarity and molality of the solution, respectively, are

Options

A. `4.5 mol L^(-1) and 5.56 mol kg^(-1)`
B. `4.5 mol L^(-1) and 4.5 mol kg^(-1)`
C. `5.56 mol L^(-1) and 4.5 mol kg^(-1)`
D. `5.56 mol L^(-1) and 5.56 mol kg^(-1)`

Answer

A

Explanation

Mass of 1L solution, `m= rho V =(1.25 g mL^(-1)) (1000mL)=1250g`
Mass of water in solution, `m_(1)= m-m_(2)= (1250-441) g 809 g= 0.809kg`
Amount of `H_(2)SO_(4)` in solution, `n_(2)= (m_(2))/(M_(2))= (441g)/(98 g mol^(-1))=4.5` mol
`m= (n_(2))/(m_(1))= (4.5 mol)/(0.809kg)= 5.56 mol kg^(-1)`