wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Claculate the mass of 5000 ml of NH3 gas at STP.

Open in App
Solution

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 = Volume at STP (litres)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 = 5.022.4

Number of moles = 0.22

Also, number of moles = Given massMolar 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 (NH3) = 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

The mass of 5000 ml of NH3 gas at STP is 3.74 g [1 Mark]


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mass Volume and Moles Relationship
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon