Question

To prepare 0.1 M $KMn{{O}_{4}}$ solution in a 250mL flask, the weight of $KMn{{O}_{4}}$ required is:
A. 4.80 g
B. 3.95 g
C. 39.5 g
D. 0.48 g

Answer 1

The number of moles of solute present per unit volume of the solution is called the Molarity of that solution. On knowing the values of molarity and volume, we can determine the amount of solute required to make a given solution by using the formula of molarity and rearranging the values in it accordingly which is given by, $Molarity=\dfrac{W}{m}\times \dfrac{1000}{V}$.

Complete Solution:
First, let us calculate the molar mass (m) of $KMn{{O}_{4}}$.
The Molecular weight of K is 40. That of Mn is 54 and Oxygen has molecular weight of 16. Here,.
{{O}_{4}} = $16 X 4 = 64$
So, total – $40 + 54 + 64 = 158 g$
We have been given that, V = 250 ml, m = 158 g, M = 0.1M
The formula to calculate molarity of a given solution is - $Molarity=\dfrac{W}{m}\times \dfrac{1000}{V}$
On rearranging this equation for W i.e. weight required we get,
$W=\dfrac{Molarity\times m\times V}{1000}$
Putting the given values in this equation,
$W=\dfrac{0.1\times 158\times 250}{1000}$
$\Rightarrow W=3.95g$
So, 3.95 g of $KMn{{O}_{4}}$ is required to make a solution of 0.1 M in a flask of 250 ml.

So, the correct answer is “Option B”.

Additional information: Molarity of a given solution is basically a measure of its concentration. The most commonly used units to define concentration is moles per litres.
- If we are given a solution which is prepared by adding several different solutions and if we know the molarity of each, by adding all the molarities we can find the cumulative molarity of the resulting solution. In an ionic solution, ionic strength is proportional to the sum of the molar concentration of salts.

Note: Another method to solve this problem is by sing the formula $c=\dfrac{n}{V}$, where n = number of moles of solute, V = volume of the solution and c = molarity of the solution.