The volume occupied by 7.23 x 10^23 molecules of carbon dioxide and 3.01 x 10^23 molecules
of Argon at 0°C and 1 atm pressure is
(A) 38 mL
(B) 3.80 L (C) 3.8 x 10^4mL
(D) 3.8 x 10^3 mL
Answers
Answer:
option c
Explanation:
carbon dioxide =26.99 L
argon =11.01L
carbon dioxide +argon =38L=3.8*(10^4)L
Given:
No. of molecules of Carbon dioxide = 7.23 × 10²³
No. of molecules of Argon = 3.01 × 10²³
To find:
Volume occupied by Carbon dioxide and Argon at 0°C and 1 atm
Solution:
We will find the no. of moles of a gas by the following formula,
where
Avogadro's Number = 6.022 × 10²³
Let "n1" be the no. of moles of carbon dioxide and "n2" be the no. of moles of argon.
∴ n1 = [7.23 × 10²³] / [6.022 × 10²³] = 1.2 moles
and
n2 = [3.01 × 10²³] / [6.022 × 10²³] = 0.499 moles
We know that at STP conditions, where the temperature is 273.15 K or 0°C and the pressure is 10⁵ Pa or 1 atm, one mole of an ideal gas has a volume of 22.4 liters.
Therefore, we will get,
The volume of 1.2 moles of Carbon dioxide = 22.4 × 1.2 = 26.88 L
and
The volume of 0.499 moles of Argon = 22.4 × 0.499 = 11.17 L
Now,
The volume occupied by the given molecules of Carbon dioxide and Argon
will be ,
= [volume of 1.2 moles of Carbon dioxide] + [volume of 0.499 moles of Argon]
= 26.88 + 11.17
= 38.05 L
≈ 38 Litres
∵ 1 L = 1000 ml
= 3.8 × 10⁴ ml ← option (C)
Thus, the volume occupied by 7.23 x 10²³ molecules of carbon dioxide and 3.01 x 10²³ molecules of Argon at 0°C and 1 atm pressure is 3.8 × 10⁴ ml.
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