`29.2% (w//w) HCl` stock, solution has a density of `1.25 g mL^(-1)`. The molecular weight of `HCl` is `36.5 g mol^(-1)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4 M HCl` is :
Correct Answer – 8
29.2% (w/w) HCl has density `=1.25` g/mL
Now, mole of HCl required in 0.4 M HCl
`=0.4xx0.2` mole `=0.08` mole
If x mol of original HCl solution is taken then mass of solution `=1.25 x`
mass of `HCl=(1.25x xx0.292)`
mole of `HCl=(1.25x xx0.292)/36.5=0.08`
So, `x=(36.5xx0.08)/(0.29xx1.25)` mol `=8` mL.
Correct Answer – 8
Molarity `(M_(1))`of stock concentrated solution can be calculated using following relation :
`M_(1)(x xx d xx10)/(m_(B))`
`=(29.2xx1.25xx10)/(36.5)=10`
`M_(1)V_(1)` (Concentrated) `= M_(2)V_(2)` (Diluted)
`10xxV_(1)=0.4xx200`
`V_(1)=8mL`]