Question

If we have 100 mL of 0.1 M $KCl$ solution. To make it 0.2 M:
A.evaporate 50 mL water
B.evaporate 50 mL solution
C.add 0.1 mol $KCl$
D.add 0.01 mol $KCl$

Answer 1

To make the required solution, first, find the volume of solution to which it is to be reduced to. Then, use the formula of molarity to evaporate that solution. Then, you will get the number of moles which is to be increased.

Formula Used:
Molarity $= \dfrac{n}{V}$
where, $n$ is the number of moles of solute, and
$V$ is the volume of solution

Complete Step by Step Solution:
For this question to be solved we have to use the formula of molarity which can be defined as the number of moles of solute dissolved per liter of solution. Mathematically, it can be represented as –
Molarity $= \dfrac{n}{V} \cdots \left( 1 \right)$
where, $n$ is the number of moles of solute, and
$V$ is the volume of solution
Now, according to the question, it is given that –
${M_1} = 0.1M \\\ {V_1} = 100mL \\\ {M_2} = 0.2M \\\$
As we know, moles remain the same upon the process of dilution. Therefore, -
${M_1}{V_1} = {M_2}{V_2}$
Putting the values in the above expression from the question, we get –
$\Rightarrow 100 \times 0.1 = x \times 0.2 \\\ \therefore x = \dfrac{{100 \times 0.1}}{{0.2}} \\\ \Rightarrow x = 50mL \\\$
So, from the above we can see that to make the required solution, we have to reduce the volume of solution to 50 mL.
Hence, this can be done when 50 mL of water or solution is evaporated. Therefore, use the formula of molarity to evaporate it.
From equation (1), we get –
Molarity $= \dfrac{n}{V}$
Putting the values in the above formula of molarity, we get –
$\Rightarrow Molarity = \dfrac{n}{{0.1}} \\\ \Rightarrow 0.2M = \dfrac{n}{{0.1}} \\\ \Rightarrow n = 0.02M \\\$
So, the number of moles should be increased from 0.01 to 0.02 by adding the 0.01 moles of $KCl$.
Hence, to make the solution of $KCl$ to 0.2 M we have to evaporate the 50 mL water and 50 mL solution and add 0.01 mol $KCl$.

Therefore, the correct options are (A), (B) and (D).

Note: If a solution having molarity ${M_1}$​ and volume ${V_1}$ ​is diluted to volume ${V_2}$ so that the new molarity is ${M_2}$ ​ then as the total number of moles will remain constant.