Question

The normality of a 26% mass/volume solution of ammonia (density 0.85 g/mL) is approximately-
A. 1.5
B. 4.0
C. 0.4
D. 15.3

Answer 1

Hint: We will first start the solution by finding out the gram equivalents of $N{H_3}$ and then convert the ml unit of the volume of the solution into L. Then we will find out the normality of the solution. Refer to the solution below for further understanding.

Complete step-by-step answer:
Formula used: number of gram equivalents$ = \dfrac{{Weight}}{{EqWeight}}$, $Normality = \dfrac{{gmEquivalent}}{{volume}}$
As we already know that the equivalent weight of $N{H_3} = 17$
as per given in the question, the concentration of $N{H_3}$ solution is 26%. This means that 100ml of $N{H_3}$ solution will definitely have 26g of $N{H_3}$.
As we know that the formula for the number of gram equivalent is weight divided by the equivalent weight. Thus-
The number of gram equivalents of $N{H_3} = \dfrac{{Weight}}{{EqWeight}}$
$ \Rightarrow \dfrac{{26}}{{17}}$ g eq
$ \Rightarrow 1.53$ g eq
The volume of the solution given is- 100ml
$
   \Rightarrow \dfrac{{100mL}}{{1000mL/L}} \\
    \\
   \Rightarrow 0.1L \\
$
Therefore, the normality of $N{H_3}$ will be the number of gram equivalents of $N{H_3}$ divided by the volume of solution (in L). So,
$
   \Rightarrow Normality = \dfrac{{gmEquivalent}}{{volume}} \\
    \\
   \Rightarrow \dfrac{{1.53gEq}}{{0.1L}} \\
    \\
   \Rightarrow 15.3N \\
$
Hence, option D is the correct option.

Note: Another definition used to calculate the concentration of a substance is the normality of chemistry. This is called 'N' and often is considered a solution's relative concentration. This is primarily found in a solution and during titration or in cases including acid-base chemistry as an indicator of reactive species. Normality is defined as the number of grams of solution equivalents in one litre of the solution, according to the standard definition. The amount of reactive mole in a product is what we claim equal.