Question

We have 100 ml of 0.1 m kcl solution. to make it 0.2m what to do ?

Answer 1

Hello mate.

We know moles remain same upon dilution: 
                                                M₁V₁=M₂V₂
We know M₁= 0.1 M
                V₁= 100ml
                M₂= 0.2M
               V₂ we have to find out(let it be x ml):
Putting in values:
0.1·100=0.2·x
10=0.2x
x= 50 ml

So we need to half the amount of solvent here. But that isn't possible since it is a solution.
∵, To make it 0.2 M we need to add KCl into it.

0.1M KCl solⁿ means = 0.1Mol/ 1 litre solution

0.2M KCl solⁿ means = 0.2 Mol/ 1 litre solution

0.2 mol of KCl= Mass of KCl required { Let required mass be y grams}
                         = (39+35.5)ₓ 0.2
                         = 14.9 g
So, we need to add 14.2 grams of KCl in a solution to make it 0.2 M.
But, in our solution, already there is 0.1 M of KCl present i.e there is already 7.45 g of KCl present. So we need to add only (14.9g- 7.45g= 7.45g).

FINAL ANSWER: WE NEED TO ADD 7.45 g OF KCl IN SOLUTION


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Answer 2

We know moles remain same upon dilution:

                                               M₁V₁=M₂V₂

We know M₁= 0.1 M

               V₁= 100ml

               M₂= 0.2M

              V₂ we have to find out(let it be x ml):

Putting in values:

0.1·100=0.2·x

10=0.2x

x= 50 ml

So we need to half the amount of solvent here. But that isn't possible since it is a solution.

∵, To make it 0.2 M we need to add KCl into it.

0.1M KCl solⁿ means = 0.1Mol/ 1 litre solution

0.2M KCl solⁿ means = 0.2 Mol/ 1 litre solution

0.2 mol of KCl= Mass of KCl required { Let required mass be y grams}

                        = (39+35.5)ₓ 0.2

                        = 14.9 g

So, we need to add 14.2 grams of KCl in a solution to make it 0.2 M.

But, in our solution, already there is 0.1 M of KCl present i.e there is already 7.45 g of KCl present. So we need to add only (14.9g- 7.45g= 7.45g).

FINAL ANSWER: WE NEED TO ADD 7.45 g OF KCl IN SOLUTION