Question

Claculate the mass of 5000 ml of $N{H}_3$ gas at STP.

Answer 1

Given, Volume of ammonia at STP = 5000 mL= 5 L

Molar volume at STP (litres) = 22.4 L

In case of gases at STP,

Number of moles = $\frac{\text{Volume at STP (litres)}}{\text{Molar volume at STP (litres)}}$ .... (i)
[1 Mark]

Putting the values of the volume of ammonia at STP and molar volume at STP (litres) in equation (i) we have:

Number of moles = $\frac{5.0}{22.4}$

Number of moles = 0.22

Also, number of moles = $\frac{\text{Given mass}}{\text{Molar Mass}}$ .... (ii)

Therefore, given mass = Number of moles x Molar mass ....(iii)

Molar mass of ammonia = Gram atomic mass of Nitrogen + (3 x Gram atomic mass of Hydrogen)
[1 Mark]

Molar mass of ammonia ($N{H}_3$) = 14 + 3 = 17 g

Substituting the values of number of moles and molar mass of ammonia in equation (iii) we get:

Mass of ammonia = 0.22 x 17 = 3.74 g

$\therefore$ The mass of 5000 ml of $N{H}_3$ gas at STP is 3.74 g [1 Mark]