Question

29.2% HCL stock solution has a density 1.25g/ml the MM of HCL is 36.5 . THe volume (Ml) of the stock solution required to prepare 200Ml of solution of 0.4M HCl is?

Answer 1

29.2% is mass percentage?

Answer 2

Yes (W/W)

Answer 3

0.4M = 0.4mol/dm^3

200 ml = 0.2 dm^3

moles of HCL in required solution is
\[0.4 \times 0.2 \text{ }mol \text{ } = 0.08 \text{ }mol\]
convert to mass as mol = mass/Mr

\[mass = 36.5 \times 0.08 =2.92g\]
now we just need to work out how much stock solution is 2.92g

Answer 4

2.92g = 29.2%

\[2.92g =0.292x\]
where x is the mass of the whole stock solution we need to add.

so
\[x = \frac{2.92}{0.292} = 10g\]

now we need to use the density of the stock solution to work out the olume containing 10g

does this make sense to you so far?

Answer 5

volume*

Answer 6

ok @eigenschmeigen plz go on

Answer 7

@eigenschmeigen the answer should be 8

Answer 8

we need 10g
with
1.25g/ml

density = mass/volume
volume = mass/density
volume = 10/1.25 = 8ml