Calculate the density of `H_(2)SO_(4)` solution if its molalilty and molarity are 94.5 and 11.5 resoectuvely.

Explanation

Calculation of mass of `H_(2)SO_(4)` in 11.5 M solution
`" Molarity"=("Moles of "H_(2)SO_(4))/("Volume of solution in litres")`
`(11.5 molL^(-1))=("Moles of" H_(2)SO_(4))/(1L)=11.5` mol
`" Moles of "H_(2)SO_(4)=(11.5mol)xx(98gmol^(-1))=1127 g`
`(" Calculation of density of the solution")/("Let the density of solution") =dg mL^(1)`
`"Mass of 1000 mL of solution"=dxxV=(1000mL)xx(d g mL^(-1))=(1000d)g`
`"Mass of solute "(H_(2)SO_(4))=1127 g`
`" Mass of solvent(water)"=((1000d-1127))/(1000)kg `
`"Molality of solution(m)"=94.5m=94.5 mol/kg`
Dencity can be calculated as follows :
`m=("Mass of" H_(2)SO_(4)//"gram molar mass")/("Mass of solvent in kg") `
` (94.5 mol kg^(1))=((1127g)//(98g mol^(-1)))/(((1000d-1127))/1000kg)`
` (94.5 mol kg^(1))=((11.5 mol))/(((1000d-1127))/1000kg)`
` (94.5 mol kg^(1))xx((1000d-1127))/(1000)kg=(11.5 mol)`
` (1000d-1127)=(11.5xx1000)/(94.5)=121.7`
`d=(121.7+1127)/1000=(1248.7)/1000=1.2487`
`"Density of solution"=1.2487 g mL^(-1)`.