An aqueous solution of `H_(2)SP_(4)` has density 1.84 g/mL. Solution containso 98% `H_(2)SO_(4)` by mass. Calculate :
(i) molarity of solution
(ii) overall molarity of solution
(iii) molar volume
(iv) specific volume
(v) relative, decrease in vapour pressure with respect to water, assuming `H_(2)SO_(4)` as non-electrolyte at his high concentration.

Explanation

(i) `M=(x xxd xx10)/(m_(B))`
`=(98xx1.84xx10)/(98)=18.4`
(ii) Weight of solution (1 litre) `= 1000 xx 1.84 = 1840` g
`w_(B)` (weight of `H_(2)SO_(4)`) `= (98)/(100)xx1840=1803.2g`
`w_(A)` (weight of water) `=1840-1803.2=36.8g`
Total number of moles `= (w_(A))/(m_(A))+(w_(B))/(m_(B))=(26.8)/(18)+(1803.2)/(98)`
`=2.044+18.4=20.444`
Overall molarity `=20.444`
(iii) Molar volume `= ("Volume")/("Total moles")=(1000)/(20.444)`
`=48.914mL//mol`
(iv) Specific volume `= (1)/("density")=(1)/(1.84)mL//g=0.543mL//g`
(v) Molality (m) `=(w_(B)xx1000)/(m_(B)xxw_(A))=(98xx1000)/(98xx2)=500`
`m=(x_(B)xx1000)/((1-x_(B))m_(A))`
`500=(x_(B)xx1000)/((1-x_(B))xx18)`
`x_(B)=0.9`
According to Raoult's law : `(Deltap)/(p_(0))=x_(B)=0.9`]