Question

The density of $N{H_4}OH$ solution is $0.6g \cdot m{L^{ - 1}}$. It contains 34% by weight of $N{H_4}OH$. Calculate the normality of the solution.
(A) 4.8 N
(B) 10 N
(C) 0.5 N
(D) 5.8 N

Answer 1

The formula to find the normality of the solution is as below.
$Normality = \dfrac{{{\text{Weight of solute(gm)}}}}{{{\text{Equivalent weight of Solute}} \times {\text{Volume of the solution(L)}}}}$
The acidity of ammonium hydroxide is one.

Complete step by step solution:
- We need to find the normality of the given solution of ammonium hydroxide. Below is the formula to find the normality of the solution.
$Normality = \dfrac{{{\text{Weight of solute(gm)}}}}{{{\text{Equivalent weight of Solute}} \times {\text{Volume of the solution(L)}}}}$
- We are not given the weight and the volume of the solution in the question directly. So, we will need to find the weight of solute which is $N{H_4}OH$ and the volume of the solution.
- It is given in the question that the solution contains 34% by weight of $N{H_4}OH$.
Suppose if the solution is of 100 gm, then the weight of $N{H_4}OH$ will be 34 gm.
We are given that the density of the solution is $0.6g \cdot m{L^{ - 1}}$. So, we can write that
${\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}$
Putting the available values in the above equation, we get
$0.6 = \dfrac{{100}}{{{\text{Volume}}}}$
So,
${\text{Volume = 166}}{\text{.66mL = 0}}{\text{.166L}}$ as 1000mL= 1L
Now, equivalent weight of solute = $\dfrac{{{\text{Molecular weight of solute}}}}{{{\text{Acidity}}}}$
So, Molecular weight = Atomic weight of N + 5(Atomic weight of H) + Atomic weight of O
$Molecular$ $weight =$ $14$ $+$ $5(1)$ $+$ $16$ $= 35$ $gmmo{l^{ - 1}}$
The acidity of $N{H_4}OH$ is one as it gives one $O{H^ - }$ ion.
So, Equivalent weight of solute = $\dfrac{{{\text{Molecular weight of solute}}}}{{{\text{Acidity}}}} = \dfrac{{35}}{1} = 35gmmo{l^{ - 1}}$
Now, we can put all the available values into the equation of normality and we will get
$Normality = \dfrac{{{\text{Weight of solute(gm)}}}}{{{\text{Equivalent weight of Solute}} \times {\text{Volume of the solution(L)}}}}$
${\text{Normality = }}\dfrac{{34}}{{35 \times 0.166}} = 5.85N$
Thus, the normality of the given solution is 5.85 N.

Therefore the correct answer to this question is (D).

Note: Do not forget that you need to put the volume of the solution in Liters if you are not putting 1000 in the numerator. Remember that we always put the weight of the solute in grams even though it is not the SI unit of mass.