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Statistics Help

**Question: **You need to prepare 100mL of 0.4M solution of KCl. You have 50mL of a 0.5 M KCl solution and 25mL of a 5% (w/v) KCl solution. What volume of % KCl and water needs to be added to the 50mL of 0.5 M KCl solution to make up 100 mL of 0.4 M KCl.

Molecular weight KCl = 74.5.

% (weight/volume) means mass (in G) in volume (100mL) Edit

**Answer: **Now we have to find out the amount of KCl in gm for each solution.

We have to prepare 100 ml 0.4 M KCl solution

now as per defination, I M 1000 ml KCl solution contains 1 mol of KCl

then 0.4 M 100 ml KCl solution contains 0.04 mol of KCl

1 mol KCl= 74.5 gm

0.04 mol KCl = 2.98 gm

That means to prepare 100 ml 0.4 M KCl solution, we need 2.98 gm KCl in 100 ml solution

In the same way 50mL of a 0.5 M KCl solution contains 1.86 gm of KCl

Now 25mL of a 5% (w/v) KCl solution means 100 ml KCl solution contains 5 gm KCl

then 25mL of a 5% (w/v) KCl solution contains 1.25 gm KCl

If we use 50mL of a 0.5 M KCl solution completely then we required (2.98-1.86)= 1.12 gm KCl from 25mL of a 5% (w/v) KCl solution

1.25 gm KCl was in 25 ml 25mL of a 5% (w/v) KCl solution

then 1.12 gm KCl will be in 22.4mL of 5% (w/v) KCl solution

Then to prepare 100mL of 0.4M solution of KCl we will use 50mL of a 0.5 M KCl solution completely and 22.4mL of 5% (w/v) KCl solution

Then water has to be added 100-50-22.4= 27.6 ml Edit

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