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Question: \[10{\text{ }}ml\] of \[0.2{\text{ }}N\;HCl\] and \[30{\text{ }}ml\] of \[0.1{\text{ }}N\;HCl\;\] to...

10 ml10{\text{ }}ml of 0.2 N  HCl0.2{\text{ }}N\;HCl and 30 ml30{\text{ }}ml of 0.1 N  HCl  0.1{\text{ }}N\;HCl\; together exactly neutralize 40 ml40{\text{ }}ml of a solution of NaOHNaOH, which is also exactly neutralized by a solution in water of   0.61 g\;0.61{\text{ }}g of an organic acid. The equivalent weight of the organic acid is:

A. 61 B. 91.5 C. 122 D. 183 {A.{\text{ }}61} \\\ {B.{\text{ }}91.5} \\\ {C.{\text{ }}122} \\\ {D.{\text{ }}183}
Explanation

Solution

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 = Normality×volumeEquivalent{\text{ }} = {\text{ }}Normality \times volume
Number of equivalent =given massequivalent weight= \dfrac{{given{\text{ }}mass}}{{equivalent{\text{ }}weight}}

Complete answer:
In the question, they given is
10  ml of  0.2  N HCl +  30  ml of  0.1  N  HCl=40  ml  of NaOH10\;ml{\text{ }}of\;0.2\;N{\text{ }}HCl{\text{ }} + \;30\;ml{\text{ }}of\;0.1\;N\;HCl = 40\;ml\;of{\text{ }}NaOH
And 40 ml of NaOH = 0.61 gm of an organic acid40{\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 HClHCl will be exactly equal to the number of milliequivalent of NaOHNaOH. And so, the number of milliequivalent of NaOHNaOH will be exactly equal to the number of milliequivalent of organic acid.
milliequivalent of HClHCl = milliequivalent of NaOHNaOH = milliequivalent of organic acid
∴ milliequivalent of HClHCl = milliequivalent of organic acid
Now, we will calculate milliequivalent of the mixture of HClHCl,
Equivalent = Normality×volumeEquivalent{\text{ }} = {\text{ }}Normality \times volume
∴ milliequivalent of HClHCl = (10 ml×0.2 N HCl)+(30 ml×0.1 N HCl)(10{\text{ }}ml \times 0.2{\text{ }}N{\text{ }}HCl) + (30{\text{ }}ml \times 0.1{\text{ }}N{\text{ }}HCl)
∴ milliequivalent of HClHCl = 2 + 32{\text{ }} + {\text{ }}3
∴ milliequivalent of HClHCl = milliequivalent of organic acid = 5
∴ number of equivalent ofHClHCl = number of equivalent of organic acid=5×10−35 \times {10^{ - 3}}

Now,
Number of equivalent of organic acid = given massequivalent weight\dfrac{{{\text{given mass}}}}{{{\text{equivalent weight}}}}
by substituting the values , we get
5×10−3=0.61gmequivalent weight5 \times {10^{ - 3}} = \dfrac{{0.61gm}}{{{\text{equivalent weight}}}}
∴ Equivalent weight = 0.615×1000\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.