Question

What amount of water should be added to 500 ml of 0.5 M HCl to make it 0.4 M HCI.​

Answer 1

Explanation:

0.5=moles/500/1000

moles=0.5×500/1000=0.25mole

0.4=0.25/V

V=0.25/0.4=0.625L=625ml

625-500=125ml or 0.125L

second method

M1V1=M2V2

0.5×500/1000=0.4×V2

V2=M1V1/M2=0.5×0.5/0.4=0.625L=625ml

amount of water added =625-500=125ml=0.125L

Answer 2

Answer:

$0.125\ L$ of water should be added to make $0.4 M\ HCL$.

Explanation:

$M_{1} = 0.5\\V_{1} =500\\M_{2} = 0.4\\V_{2} = ?$

∵ The amount of water should be added to the make $HCL$ = $V_{2}$

hence

  ⇒         $M_{2} V_{2} =M_{2} V_{2}$           $(I)$

now put the value on the given equation $(I)$, we get

  ⇒         $\frac{0.5\ * \ 500}{1000} =0.4\ * V_{2}$

  ⇒         $V_{2} = 0.625\ L\\V_{2} = 625\ ml$

So

   now the rest of the volume should be subtracted from the  $V_{2}$, we get

⇒

         $V_{2} = 625-500\ ml\\V_{2} = 125\ ml\\V_{2} = 0.125\ L$

Therefore,

                 $0.125\ L$ of water should be added to make $0.4 M\ HCL$.

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