Question

In a volumetric experiment, it was found that a solution of $KMn{{O}_{4}}$ is reduced to $MnS{{O}_{4}}$. If the normality of a solution is 1N, then molarity of solution will be:
(A) 0.5M
(B) 0.2M
(C) 1M
(D) 0.4M

Answer 1

Hint: Try to recall that normality is a measure of concentration equal to gram equivalent weight per litre of solution and normality is the product of molarity of the solution and valency factor. Now, by using this you can easily find the correct option from the given ones.

Complete step by step solution:
-We know that normality in chemistry is one of the concentration terms like molarity or molality and it is formulated as the number of gram equivalent of solute divided by the volume of solution in litres. The formula for this is:
\[N\text{=}\dfrac{\text{no}\text{. of gram equivalents}}{\text{volume of solution in liters}}\]
-A gram equivalent of a solute means given mass of solute ($w$) divided by equivalent mass and equivalent mass of a substance is equal to molar mass ($M$) divided by valency factor ($n$). The formulae involved are:
\[\begin{align}
& \text{no}\text{. of gram equivalents}=\dfrac{w}{\text{equivalent mass}} \\
& \text{equivalent mass}=\dfrac{M}{n} \\
\end{align}\]
Putting the value of equivalent mass in the formula for the number of gram equivalents, we get:
\[\text{no}\text{. of gram equivalents}=\dfrac{w\times n}{M}\]
-So, if we put the formula of equivalent mass and gram equivalent to the formula of normality we would get:
\[N=\dfrac{w\times n}{M\times V}\]
By the above formula, we can relate normality with molarity,
Number of moles of solute, $m=\dfrac{w}{M}$
Hence, the formula for normality becomes:
\[N=\dfrac{m\times n}{V}\]
Now we know that the number of moles of a substance divided by the volume is equal to the molarity, which is represented by:
\[\text{molarity}=\dfrac{m}{V}\]
Substituting this in the equation, we get:
\[\begin{align}
 & N=M\times n \\
 & normality=molarity\times n \\
\end{align}\]
-Calculation:
Valency factor ($n$) is the number of electrons involved in reaction in which manganese changes from +7 to +2.
\[M{{n}^{+7}}+5{{e}^{-}}\to M{{n}^{+2}}\]
So, the above reaction valency factor ($n$) is 5.
Given, normality of solution (N) = 1N
Now, using the formula relating molarity and normality, we get:
\[\begin{align}
& normality=molarity\times n \\
& molarity=\dfrac{normality}{n} \\
& molarity=\dfrac{1N}{5} \\
& molarity=0.2M \\
\end{align}\]

Therefore, from above we can conclude that option B is the correct option to the given question.

Note: Note that normality is a property of the mixture, and will vary with the use of more or less dissolving liquid to place the substance of interest into a solution. Also, you should remember that in acid-base chemistry, normality is used to express the concentration of hydronium ions or hydroxides ions in a solution.