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Sagot :
Answer: The volume of given container is 3.83 L.
Explanation:
Given: Mass of [tex]H_{2}[/tex] = 0.224 g
Mass of [tex]N_{2}[/tex] = 1.06 g
Mass of Ar = 0.834 g
Since, moles is the mass of a substance divided by its molar mass. Therefore, moles of given substances present in the mixture are as follows.
Moles of [tex]H_{2}[/tex] are:
[tex]Moles = \frac{mass}{molar mass}\\= \frac{0.224 g}{2 g/mol}\\= 0.112 mol[/tex]
Moles of [tex]N_{2}[/tex] are:
[tex]Moles = \frac{mass}{molar mass}\\= \frac{1.06 g}{28 g/mol}\\= 0.038 mol[/tex]
Moles of Ar are:
[tex]Moles = \frac{mass}{molar mass}\\= \frac{0.834 g}{40 g/mol}\\= 0.021 mol[/tex]
Total moles = (0.112 + 0.038 + 0.021) mol = 0.171 mol
Now, using ideal gas equation the volume is calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]V = \frac{nRT}{P}\\= \frac{0.171 mol \times 0.0821 L atm/mol K \times 273 K}{1 atm}\\= 3.83 L[/tex]
Thus, we can conclude that the volume of given container is 3.83 L.
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