Question

22.4 litres of a gas weighs 70 g at S.T.P. Calculate the
weight of the gas if it occupies a volume of 20 litres at
27°C and 700 mm Hg of pressure.

Please give the answer with detailed explanation. I need it rn.​

Answer 1

Given:-

→ 22.4 L of a gas weighs 70g at STP.

→ Given volume of the gas = 20 L

→ Pressure = 700 mm Hg

→ Temperature = 27°C

To find:-

→ Weight of the gas at given conditions.

Solution:-

Firstly, let's convert the units of temperature and pressure :-

Temperature:-

⇒ 0°C = 273 K

⇒ 27°C = 273 + 27

⇒ 300 K

Pressure:-

⇒ 1 mm Hg = 1/760 atm

⇒ 700 mm Hg = 700 × 1/760

⇒ 0.92 atm

________________________________

Since 22.4 L of the gas weigh 70g at STP, thus molar mass of the gas is 70g/mol.

Now, let's calculate the number of moles of the gas by using Ideal Gas equation :-

⇒ PV = nRT

⇒ 0.92 × 20 = n × 0.082 × 300

⇒ 18.4 = 24.6n

⇒ n = 18.4/24.6

⇒ n = 0.75

________________________________

Weight = No of moles × Molar mass

= 0.75 × 70

= 52.5g

Thus, required weight of the gas is 52.5g .

Answer 2

Given :-

22.4 litres of a gas weighs 70 g at S.T.P. Calculate the

weight of the gas if it occupies a volume of 20 litres at

27°C and 700 mm Hg of pressure.

To find :-

Weight of gas

Solution :-

We know that

At first

27 C = 273 + 27 = 300 K

Now

1 mm Hg = 1/760 atom

0.92 1 mm Hg = 700 * 1/760 atm = 700/760 = 0.92

 

=> 0.92 *  20 = n × 0.082 × 300

=> 18.4 = 24.6n

n = 0.75

Now

Weight = 0.75 * 70 = 52.5 g