Question

A cylinder contains 68g of ammonia gas at stp.
1. What is the volume occupied by this gas ?
2. How many moles of ammonia are present in the cylinder?
3. How many molecules of ammonia are present in the cylinder?
N=14, h=1

Answer 1

RMM of ammonia = 14 + 1*3 = 17
No. of moles - 68/17 = 4
Volume = 22.4*4 = 89.6 LITRES
NO. of molecules = 4 * Avogadro's no. = 4 * 6.022 * 10^23

Answer 2

Ammonia gas is  N H₃ .  Molecular mass = 14 + 3 = 17 amu

Number of moles of Ammonia in 68 gms = 68/17 = 4 moles

1 mole of an ideal gas occupies 22.4 liters at STP
Hence 4 moles occupy 4 * 22.4 = 89.6 liters 

Number of molecules present in four moles of an ideal gas at any temperature and pressure = 4 * Avogadro number = 4 * 6.023 * 10^23