Question

\[10{\text{ }}ml\] of \[0.2{\text{ }}N\;HCl\] and \[30{\text{ }}ml\] of \[0.1{\text{ }}N\;HCl\;\] together exactly neutralize \[40{\text{ }}ml\] of a solution of \[NaOH\], which is also exactly neutralized by a solution in water of \[\;0.61{\text{ }}g\] of an organic acid. The equivalent weight of the organic acid is:
\[
  {A.{\text{ }}61} \\
  {B.{\text{ }}91.5} \\
  {C.{\text{ }}122} \\
  {D.{\text{ }}183}
\]

Answer 1

Hint: We have to find the equivalent weight of an unknown organic compound. Therefore, we can use equivalent formulas to solve the problem. Before that, we understand the given information.
\[Equivalent{\text{ }} = {\text{ }}Normality \times volume\]
Number of equivalent $= \dfrac{{given{\text{ }}mass}}{{equivalent{\text{ }}weight}}$

Complete answer:
In the question, they given is
\[10\;ml{\text{ }}of\;0.2\;N{\text{ }}HCl{\text{ }} + \;30\;ml{\text{ }}of\;0.1\;N\;HCl = 40\;ml\;of{\text{ }}NaOH\]
And \[40{\text{ }}ml{\text{ }}of{\text{ }}NaOH{\text{ }} = {\text{ }}0.61{\text{ }}gm{\text{ }}of{\text{ }}an{\text{ }}organic{\text{ }}acid\]
Therefore, this data suggests that,
The number of milliequivalent of Mixture of \[HCl\] will be exactly equal to the number of milliequivalent of \[NaOH\]. And so, the number of milliequivalent of \[NaOH\] will be exactly equal to the number of milliequivalent of organic acid.
milliequivalent of \[HCl\] = milliequivalent of \[NaOH\] = milliequivalent of organic acid
∴ milliequivalent of \[HCl\] = milliequivalent of organic acid
Now, we will calculate milliequivalent of the mixture of \[HCl\],
\[Equivalent{\text{ }} = {\text{ }}Normality \times volume\]
∴ milliequivalent of \[HCl\] = \[(10{\text{ }}ml \times 0.2{\text{ }}N{\text{ }}HCl) + (30{\text{ }}ml \times 0.1{\text{ }}N{\text{ }}HCl)\]
∴ milliequivalent of \[HCl\] = \[2{\text{ }} + {\text{ }}3\]
∴ milliequivalent of \[HCl\] = milliequivalent of organic acid = 5
∴ number of equivalent of\[HCl\] = number of equivalent of organic acid=\[5 \times {10^{ - 3}}\]

Now,
Number of equivalent of organic acid = $\dfrac{{{\text{given mass}}}}{{{\text{equivalent weight}}}}$
by substituting the values , we get
$5 \times {10^{ - 3}} = \dfrac{{0.61gm}}{{{\text{equivalent weight}}}}$
∴ Equivalent weight = $\dfrac{{0.61}}{5} \times 1000$
∴ Equivalent weight = 122

Hence, we can conclude that the correct option is C.

Note:
We must remember that the normality of a solution is the gram equivalent weight by liter of solution. i.e. volume. The weight of a substance in grams is always numerically equal to the equivalent weight and it is known as a gram equivalent.