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=pV `=(1.25 gmL^(-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/V=(4.5 mol)/(1L)="4.5 mol L"^(-1)`
`m=n_2/m_1=(4.5 mol)/(0.809kg)="5.56 mol kg"^(-1)`