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Answer :
The molar mass of the gas sample is 32 g/mol.
Explanation:
Using the ideal gas law, we can determine the number of molecules present in the volume given,
[tex]PV = nRT\\\\n = \frac{PV}{RT}[/tex]
Here, Pressure P is given as 1.2 atm, the volume V is given as 37.6 L and temperature is given as 275 K with the gas constant as 0.082 L atm mol⁻¹ K⁻¹.
[tex]n = \frac{1.2*37.6}{275*0.082} = \frac{45.12}{22.55} =2[/tex]
So, the 2 moles of gas is present in 64 g of gas sample. Thus, as molar mass is found by the ratio of mass of the sample to its amount of moles present in the sample, the molar mass can be found as below.
[tex]Molar mass = \frac{Mass}{No.of moles} = \frac{64}{2} = 32 g/mol[/tex]
So, the molar mass of the gas sample is 32 g/mol.
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Rewritten by : Barada
32.02 grams/mole is the molar mass of the gas.
Explanation:
Data given about the gas:
sample of the gas = 64 grams
volume of the gas = 37.6 L
temperature of the gas = 275K
pressure of the gas = 1.2 atm
R = 0.0821 atm l/mole K
molar mass=?
The data provided will be used to know the molar mass by using ideal gas law equation:
PV = nRT (n = [tex]\frac{mass}{molar mass}[/tex]
putting the values in the equation:
n = [tex]\frac{PV}{RT}[/tex]
Putting the value of n and then writing the equation:
[tex]\frac{mass}{molar mass}[/tex] = [tex]\frac{PV}{RT}[/tex]
putting the values in the equation
molar mass = [tex]\frac{64 X 0.0821 X 275}{37.6 X 1.2}[/tex]
= 32.02 grams/mole
