Question

17 g of $N{H_3}$ is present in $500.0ml$ volume. Its active mass is:
a) $1.0M{L^{ - 1}}$
b) $0.5M{L^{ - 1}}$
c) $1.5M{L^{ - 1}}$
d) $2.0M{L^{ - 1}}$

Answer 1

The active mass of a ammonia $\left( {N{H_3}} \right)$ is obtained on dividing number of moles of ammonia by the volume of solution in litres and molar mass is obtained by multiplying the subscript of nitrogen $\left( N \right)$ with molar mass of N and multiplying the subscript of hydrogen $\left( H \right)$ with molar mass of H and then adding both the values of nitrogen and hydrogen. Then, the number of moles is calculated on dividing the given mass by the molar mass of ammonia.

Complete Step by step answer:
Active mass is defined as the molar concentration of a substance which takes part in a chemical reaction. In other words, the number of moles present in one litre of a solution. It is also known as molarity. Its unit is $\dfrac{{mol}}{{litre}}$ or $M{L^{ - 1}}$.
Active mass of a substance $\left( A \right)$ = $\dfrac{{number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}the{\text{ }}substance}}{{volume{\text{ }}of{\text{ }}solution{\text{ }}in{\text{ }}litre}}$
So, now we’ll calculate the number of moles of ammonia $\left( {N{H_3}} \right)$ and mass of $N{H_3}$ is given as 17 g. Number of moles $\left( n \right)$ of $N{H_3}$ is calculated on dividing the mass of ammonia $\left( {N{H_3}} \right)$ by molecular or molar mass of $N{H_3}$ i.e.,
$n = \dfrac{m}{M}$ where m is the given mass of $N{H_3}$ and M is the molar mass of $N{H_3}$
So, now we’ll calculate the molar mass of NH3 i.e., $M = 1 \times N + 3 \times H$.
$M = 1 \times 14 + 3 \times 1$ where molar mass of nitrogen $\left( N \right)$ is 14g and molar mass of hydrogen $\left( H \right)$ is 1g.
$M = 14 + 3$
$\Rightarrow M = 17g$
Now, $n = \dfrac{m}{M}$
$n = \dfrac{{17}}{{17}}$
$\Rightarrow n = 1$
Now, Active mass of a substance $\left( A \right)$ = $\dfrac{{number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}the{\text{ }}substance}}{{volume{\text{ }}of{\text{ }}solution{\text{ }}in{\text{ }}litre}}$
$A = \dfrac{1}{{500}}$ where volume is given as 500 ml
$\Rightarrow A = \dfrac{1}{{500 \times 0.001}}$ $\left[ {1L = 0.001ml} \right]$
$\Rightarrow A = \dfrac{{1000}}{{500}}$
$\Rightarrow A = 2M{L^{ - 1}}$

Therefore, option D is correct.

Note: Remember that the active mass of a substance is known as molarity or molar concentration of the substance as we are more familiar with the term molarity and which would help us to solve this question easily and also memorize the molar mass of nitrogen and hydrogen. The volume in the formula is given in litres, so we have to convert the given volume to litre.