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29.2 % (W//W) HCl stock solution has a d...

`29.2 % (W//W) HCl` stock solution has a density of `1.25 "g mL"^(-1)`. The molecular mass of `HCl` is `36.5 g mol^(1)`. What is the volume (mL) of stock required to prepare 200 mL of solution of `0.4 M HCl` ?

Text Solution

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Mass of solution `= 100 g`
Density of solution `= 1.25 "g mL"^(-1)`
Volume of 100 g of the solution `= ("Mass of solution")/("Density of solution")`
`=((100g))/((1.25"g mL"^(-1)))=80 "mL"`
Mass of HCl in the solution = 29.2 g
Molarity of solution `(M)=("Mass of HCl/Molar mass")/("Volume of solution in litres")`
`=((29.2g)//(36.5"g mol"^(-1)))/((0.08L))=10"mol L"^(-1)=10M`
Volume of stock solution required to prepare 200 mL of 0.4 M HCl solution :
`M_(1)V_(1)-=M_(2)V_(2)`
`(10M)xxV_(1)=(0.4M)xx(200mL)`
`V_(1)=((0.4M)xx(200mL))/((10M))`.
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