Equal volumes of 30% by mass of `H_(2)SO_(4)` (density 1.218g `mL^(-1))` and of `H_(2)SO_(4)` (density 1.610 g mL^(-1)) are mixe. If the density of te

Calculation of molarity of the solution
Let V mL of each sample of `H_(2)SO_(4)` bemixed for 30% `H_(2)SO_(4)`
Mass of `H_(2)SO_(4)`=30g ,Mass of solution=100 g
`" Volume of solution"=("Mass of soulution")/("Density")= ((100g))/((1.218gmL^(-1)))=100/(1.218)=mL`
`100/1.218` mL of sample contains `H_(2)SO_(4)=30 g`
V mL of contains `H_(2)SO_(4)=((30g)xx(V mL))/((100//1.18mL))=(0.3654 V)g`
For 70% `H_(2)SO_(4)`
Mass of `H_(2)SO_(4)`=70g, Mass of solution=100 g
`"Volume of solution"=("Mass of solution")/("Density")=((100g))/((1.610gmL^(-1)))=100/(1.610)mL`
`100/(1.610)` mL of sample sontains `H_(2)SO_(4)=70g`
V mL of sample contains `H_(2)SO_(4)=((70g)(VmL))/((100//1.610mL))=(1.127V)g`
On mixing the two sample os `H_(2)SO_(4)`
Total mass of `H_(2)SO_(4)=0.3654V+1.127V=(1.127V)g`
Total volume of solution=2VmL=(0.002V)L
`"Molarity of solution(M)"= ("Total mass of" H_(2)SO_(4)//"Molar mass")/("Voluume of solution in litres") `
`=((1.4924 Vg)//(98gmol^(-1)))/((0.002V)L)=7.614 mol L^(-1)=7.614 M`
Calculation of malality of the solution.
Total mass of solution= `d xx" volume "=(1.425 gmL^(-1)xx2VmL)=(2.85v)g`
Total mass of acid =(1.4924 V)g
Mass of water = `(2.85 V-1.4924 V) g=(1.3576xx10^(-3)V) kg`
`"Molality of solutin (m)"=("Mass of" H_(2)SO_(4)//"Molar mass")/("Mass of solvent in kg")=((1.4924V)g//(98gmol^(-1)))/((1.3576xx10^(-3)V)kg)`
=11.217 mol `kg ^(-1)`=11.217 m