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)`.
Find the Volume (V)(mL) of stock solution required to prepare a 500 mL solution of 0.4 M HCl.
Report your answer as `V//5`

Answer

4

Explanation

29.2% (w/w) HCl has density=1.25g/ml
Now, mole of HCl required in 0.4 M HCl
`=0.4xx0.5` mole =0.2 mole
If v mol of orginal HCl solution is taken
then volume of solution =1.25 v
mass of HCl=(1.25 v x 0.292)
mole of HCl `=(1.25vxx0.292)/36.5=0.2`
so, `V=(36.5xx0.2)/(0.29xx1.25)` mol =20 mL
V/5=4