Question

Calculate the amount of $ {\text{KOH}} $ required to prepare $ {\text{100mL}} $ of $ {\text{0}}{\text{.1M}} $ solution. (Atomic weight of K=10, O=16, H=1).

Answer 1

Hint: In the above question, since, we have to find the amount of $ {\text{KOH}} $ required to prepare $ {\text{100mL}} $ of $ {\text{0}}{\text{.1M}} $ solution , we first have to find out number of moles of $ {\text{KOH}} $ present. Then only we can find out the amount of $ {\text{KOH}} $ required.

Formula used:
$ {\text{Molarity = }}\dfrac{{\text{n}}}{{\text{V}}} $
Where n is the number of moles of solute and V is the volume of solution present in litres.
 $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $
Where n= number of moles of the substance, m=given mass of the substance, M=molar mass of the substance.

Complete step by step solution:
We know that:
 $ {\text{Molarity = }}\dfrac{{\text{n}}}{{\text{V}}} $
Since molarity of the solution is given as $ {\text{0}}{\text{.1M}} $ and volume of solution is given as $ {\text{100mL = }}\dfrac{{100}}{{{\text{1000}}}}{\text{L}} $ .
Now, number of moles of $ {\text{KOH}} $ = Molarity $ \times $ V = $ {\text{0}}{{.1 \times }}\dfrac{{100}}{{{\text{1000}}}}{\text{ = 0}}{\text{.01}} $
Let us now find out molar mass of $ {\text{KOH}} $
Molar mass of $ {\text{KOH}} $ = atomic mass of K + atomic mass of O + atomic mass of H
Molar mass of $ {\text{KOH}} $ = $ {\text{10 + 16 + 1 = 27}} $
As we know that: $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $
By cross-multiplication, we get:
 $ {{n \times M = m}} $
Which means that:
Mass of $ {\text{KOH}} $ = number of moles of $ {\text{KOH}} $ present $ \times $ Molar mass of $ {\text{KOH}} $
Substituting the values, we get:
Mass of $ {\text{KOH}} $ = $ {\text{0}}{{.01 \times 27 = 0}}{\text{.27g}} $
 $ \therefore $ The amount of $ {\text{KOH}} $ required to prepare $ {\text{100mL}} $ of $ {\text{0}}{\text{.1M}} $ solution is $ {\text{0}}{\text{.27g}} $ .

Note:
Sometimes, there is a confusion between molarity and molality. In such a case, remember, the difference between molarity and molality lies in the denominator part of the ratio. In molarity, we divide the number of moles by the volume of the solution. And in molality, we divide the number of moles by mass of solvent in kg.
Solution=Solute + Solvent
So, in molality solute mass must be subtracted to get the correct result.
In molarity, the unit of the volume of the solution should be taken into consideration. It should be converted into litres in order to get the correct result.