Respuesta :
Answer: Option (e) is the correct answer.
Explanation:
According to Graham's law, time taken for effusion is directly proportional to square root of molar mass.
[tex]\frac{T_{N_{2}}}{T_{Gas}} = \sqrt{\frac{M_{N_{2}}}{M_{Gas}}}[/tex]
where, [tex]T_{N_{2}}[/tex] = Time taken by the nitrogen gas
[tex]T_{Gas}[/tex] = Time taken by the other gas
[tex]M_{N_{2}}[/tex] = molar mass of nitrogen
[tex]M_{Gas}[/tex] = molar mass of other gas
Therefore, we will calculate the molar mass of unknown as follows.
[tex]\frac{T_{N_{2}}}{T_{Gas}} = \sqrt{\frac{M_{N_{2}}}{M_{Gas}}}[/tex]
[tex]\frac{48 sec}{63 sec} = \sqrt{\frac{28 g/mol}{M_{Gas}}}[/tex]
Squaring on both the sides,
[tex]M_{Gas} = (\frac{63}{48})^{2} \times 28[/tex]
= [tex]\frac{3969}{2304} \times 28[/tex]
= 48.2 g/mol
Thus, we can conclude that molar mass of the given gas is 48 g/mol.
Answer:
The molar mass of the gas is 48 g/mol
Explanation:
Step 1: data given
Time to effuse : 63s
Time to effuse for nitrogen = 48s
Molar mass N2 = 28 g/mol
Step 2: Calculate the molar mass
√M1 / t1 = √M2 / t2
⇒ with M1 = the molar mass of N2 = 28 g/mol
⇒ with t1 = the time to effuse for nitrogen = 48s
⇒ with M2 = the molar mass of the gas = TO BE DETERMINED
⇒ with t2 = the time to effuse the gas = 63 s
√28 / 48 = √M2 /63
0.110 = √M2 /63
M2 = 48 g/mol
The molar mass of the gas is 48 g/mol
