22.4 litres of a gas weighs 70g at S.T.P. Calculate the weight of the gas if it occupies a volume of 20litres at 27°C and 700mm of Hg Pressure.
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Chapter: Gas Laws.
Class 9.
Answers
Given:-
→ Weight of 22.4L of the gas at
STP = 70g
→ Given temperature = 27° C
→ Volume occupied by the gas = 20L
→ Pressure = 700mm Hg
To find:-
→ Weight of the gas at given conditions.
Solution:-
Firstly, let's convert the given pressure from mm Hg to atm.
=> 1 mm Hg = 1/760 atm
=> 700 mm Hg = 700×1/760
=> 0.92 atm
Temperature :-
=> 0° C = 273 K
=> 27° C = 273 + 27
=> 300 K
Now, let's calculate the number of moles by using the Ideal gas equation .
PV = nRT
Where :-
• P is the pressure.
• V is volume of the gas.
• n is the number of moles.
• R is ideal gas constant [0.082].
• T is the temperature.
=> 0.92 × 20 = n × 0.082 × 300
=> 18.4 = 24.6n
=> n = 18.4/24.6
=> n = 0.748 mole
_______________________________
We know that 1 mole of the gas is present at STP.
∵ Weight of 1 mole = 70g
∴ Weight of 0.748 mole :-
= 0.748×70/1
= 52.36g
Thus, the required weight of the gas is 52.36g .
Answer:
refer to the above attachment
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