What volume of 95% sulphuric acid (density ="1.85 g/cm"^(3)) and what mass of water must be taken to prepare "100 cm"^(3) of 15% solution of sulphuric acid (density = "1.10 g/cm"^(3)) ?

What volume of 95% sulphuric acid (density ="1.85 g/cm"^(3)) and what

1.	What volume of 95% sulphuric acid (density ="1.85 g/cm"^(3)) and what mass of water must be taken to prepare "100 cm"^(3) of 15% solution of sulphuric acid (density = "1.10 g/cm"^(3)) ?
Answer» Solution :`95% H_(2)SO_(2)` MEANS that `" 95 g of "H_(2)SO_(4)` are present in 100 g of the solution `"Volume of 100 g of this solution"=("100 g")/("1.85 g cm"^(-1))="54.05 cm"^(3)` Molarity of `95%H_(2)SO_(4)=("95 g")/("98 g mol"^(-1))xx(1)/("54.05 cm"^(3))xx"1000 cm"^(3)L^(-1)="17.93 M` `15% H_(2)SO_(4)` means that 15 g `H_(2)SO_(4)` are present in 100 g of the solution `"Volume of 100 g of solution "=("100 g")/("1.10 g cm"^(-1))="90.01 cm"^(3)` `"Molarity of 15% "H_(2)SO_(4)=("15 g")/("98 g mol"^(-1))xx(1)/("90.01 cm"^(3))xx"1000 cm"^(3)L^(-1)="1.68 M"` `"Applying molarity equation"{:(M_(1)xxV_(1),=,M_(2)xxV_(2)),((95%H_(2)SO_(4)),,(15%H_(2)SO_(4))),(17.93xxV_(1),=,1.68xx100 or V_(1)=9.37~="9.4 cm"^(3)):}` Mass of `100cm^(3)` of `15% H_(2)SO_(4)` to be prepared `=100cm^(3)xx"1.10 g cm"^(-3)="110 g"` `"Mass of 9.4 cm"^(3)xx"1.85 g cm"^(-3)="17.4 g"` `THEREFORE"Mass of water to be taken "=110-17.4 g = 92.6 g`