Question

What volume of $0.232{\text{ N}}$ solution will contain $3.17$ milli-equivalent of solute.
$(i){\text{ 137 ml}}$
$(ii){\text{ 13}}{\text{.7 ml}}$
$(iii){\text{ 27}}{\text{.3 ml}}$
$(iv){\text{ 12}}{\text{.7 ml}}$

Answer 1

The equivalent of solute is the ratio of weight of the solute to the equivalent weight of the solute. It has no units and hence called an equivalent number of solutes. We will use the relation between the normality and equivalent of solute to find the volume of the solution.
Formula Used:
${\text{Normality = }}\dfrac{{{\text{equivalent of solute}}}}{{{\text{Volume of solution in litre}}}}$

Complete Answer:
The normality of a solute is defined as the ratio of the equivalent of solute to the volume of solution in litres. The number of equivalents of solute is defined as the ratio of weight of given solute to the equivalent weight of the solute. This can be represented as,
${\text{Number of equivalent = }}\dfrac{{{\text{weight of solute}}}}{{{\text{equivalent weight of solute}}}}$
According to the equation we are given with the normality of the solution which is equal to $0.232{\text{ N}}$. The number of milli-equivalent is $3.17$. Now we have to find the volume of solution. As we can see that we are given with milli-equivalent, therefore the volume of solution will be in milli-litre. Since we know the relation between the normality, equivalent and volume of solution we can write as,
${\text{Normality = }}\dfrac{{{\text{milli - equivalent of solute}}}}{{{\text{Volume of solution in ml}}}}$
The equation can be deduced as,
${\text{Volume of solution in ml = }}\dfrac{{{\text{milli - equivalent of solute}}}}{{{\text{Normality}}}}$
Now substituting the values we get,
${\text{Volume of solution in ml = }}\dfrac{{3.17}}{{0.232}}$
${\text{Volume of solution in ml = 13}}{\text{.66 ml}}$
Thus the volume of solution is $13.66{\text{ ml}}$ which can be rounded-off to $13.7{\text{ ml}}$. Thus the correct option is $(ii){\text{ 13}}{\text{.7 ml}}$ .

Note:
If we are given the number of equivalents then the volume of solution will be in litre. This is why the units of a given quantity must be observed clearly. We can also convert the volume in litre to volume in milli-litre. We can also convert molarity into normality by multiplying it with n-factor.