Question

How much does 9.8 liters of ammonia (NH3) weigh at standard temperature and pressure (273 K and 100 kPa)?

Answer 1

9.8 × 17 / 22.4 = 7. 44g

Answer 2

Answer: The mass of ammonia will be 7.34 g

Step-by-step explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

`PV=nRT`

Or,

`PV=(m)/(M)RT`

where,

P = pressure of the gas = 100 kPa

V = Volume of gas = 9.8 L

m = Given mass of ammonia = ? g

M = Molar mass of ammonia = 17 g/mol

R = Gas constant =
`8.31\text{L kPa }mol^(-1)K^(-1)`

T = temperature of the gas = 273 K

Putting values in above equation, we get:

`100kPa* 9.8L=(m)/(17g/mol)* 8.31\text{L kPa }mol^(-1)K^(-1)* 273K\\\\m=7.34g`

Hence, the mass of ammonia will be 7.34 g