Correct Answer - 1.68M
3.36M
1.8 M
Calcultion of molality of solution
Mass os `H_(2)SO_(4)` in solution= 15 g
Mass os solvent (water) = 100-5 = 85 g
Molar mass of `H_(2)SO_(4) = 98 " g mol"^(-1)`
`"Molality of solution (m)"=("Mass of" H_(2)SO_(4)//"Molar mass of" H_(2)SO_(4))/("Mass of water in kg")`
`=(15g//(98" g mol"^(-1)))/((85//1000 kg))=1.8 m`
Step II. Calcultation of molarity of solution
`"Density of solution"=1.10" g mol"^(-1)`
`"Volume of 100 g of solution"=("Mass")/("Density")((100g))/((1.10" g mL"^(-1)))=90.9 mL`
`"Molarity of solution (M)"=("Mass of" H_(2)SO_(4)//"Molar mass of" H_(2)SO_(4))/("Volume of solution in litres")`
`((15g)//(98" g mol"^(-1)))/((90.9//1000 L))=1.68 M`.
Step III. Calculatation of normallity of solution
Normality of solution = `"Molarity "xx "Basicity"=(1.68xx2)=3.36 N`.
