Step I. Calculation of molarity of the solution
Mass of `H_(2)SO_(4)=93` g
Molar mass of `H_(2)SO_(4)=2xx1+32 +4 xx 16 = 98 u =(98 "g mol"^(-1))`
Volume of solution `= 100 mL = (100)/(1000)=0.1 L`
Molarity of the solution `(M)=("Mass of "H_(2)SO_(4)//"Molar mass")/("Volume of solution in litres")=((93g)//(98"g mol"^(-1)))/((0.1 L))`
`9.49"mol L"^(-1)=9.49 M`
Step II. Calculation of molality of the solution
Mass of 100 mL of solution `= d xx V = (1.84 "g mL"^(-1))xx(100 mL)=184 g`
Mass of `H_(2)SO_(4)=93 g`
Mass of water in the solution `= 184-93 = 91 g = (91)/(1000)=0.091 kg`
Molality of the solution `(m)=("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")=((93g)//(98"g mol"^(-1)))/((0.091kg))`
`=10.43"mol kg"^(-1)=10.43 m`.
