What volume of 95% sulphuric acid (density = 1.85 g mL^(-1)) and what mass of water must be taken to prepare 100 mL of 15% solution of sulphuric acid (density = 1.1 g mL^(-1)) ?

What volume of 95% sulphuric acid (density = 1.85 g mL^(-1)) and what

1.	What volume of 95% sulphuric acid (density = 1.85 g mL^(-1)) and what mass of water must be taken to prepare 100 mL of 15% solution of sulphuric acid (density = 1.1 g mL^(-1)) ?
Answer» Solution :Step. Calculation of molarity of 95% `H_(2)SO_(4)` solution. `"Mass of "H_(2)SO_(4)" in 100 g of solution"=95 g` `"Density of solution"=1.85 ML^(-1)` `"Volume of 100 g of solution"=("Mass")/("Density")=((100g))/((1.85"g mL"^(-1)))=54.05 mL` `"Molarity (M)"=("Mass of "H_(2)SO_(4)"/Molar mass")/("Volume of solution in LITRES")=((95G)//(98"g mol"^(-1)))/(0.05405 L)=17.93 M` Step II. Calculation of molarity of 15% `H_(2)SO_(4)` solution. `"Mass of "H_(2)SO_(4)" in 100g of solution"=15g` `"Density of solution"=1.1" g mL"^(-1)` `"Volume of 100g of the solution"=("Mass")/("Density")=((100g))/((1.1"g mL"^(-1)))=90.90 mL` `"Molarity (M)"=("Mass of "H_(2)SO_(4)"/Molar mass")/("Volume of soltuion in litres")=((15g)//98"g mol"^(-1))/((0.0909L))=1.68 M` Step III.Calculation of volume of 17.93 M solution required. The volume can be calculated with the help of molarity equation, `M_(1)V_(1)=M_(2)V_(2)` `(17.93 M )xx(V_(1))=(1.68 M)xx(100mL)` `V_(1)=((1.68M)xx(100mL))/((17.93M))=9.37 mL.` Step IV. Calculation of mass of WATER to be taken. `"Mass of 100 mL of 15% "H_(2)SO_(4)" to be prepared" =Vxx=d=(100 mLxx1.1"g mL"^(-1))=110g` `"Mass of 9.37 mL of 95% "H_(2)SO_(4)=Vxxd=(9.37 mLxx1.85"g mL^(-1))=17.33 g` `"Mass of water to be taken"=(110-17.33)=92.67 g`