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

9.37 mL; 92.67 g

Explanation

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`