Question

100cm³ of ethane gas diffuses through a porous plug in 100 seconds.What is the molecular mass of the gas Q if 100cm³ of the gas diffuses through the same plug in 121 secknds under the same condition?(C=12.0,H=1.0)

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Answer 1

The molar mass of gas Q is 43.923 g/mol

Step-by-step explanation:

The given volume of ethane gas that diffuses through a porous plug in 100 seconds = 100 cm³

Therefore;

The rate of diffusion of ethane gas through the porous plug,
`v_(ethane)`, is given as follows;

`v_(ethane)` = (100 cm³/100 s) = 1 cm³/s

The molar mass of ethane, C₂H₆ = 2×12 g/mol + 6×1 g/mol = 30 g/mol

The given volume of gas, Q, that diffuses through a porous plug in 121 seconds = 100 cm³

∴ The rate of diffusion of the gas, Q,
`v_Q` = 100/121 cm³/s

Graham's Law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of the molecular mass of the gas

Mathematically, we have;

`(v_A)/(v_B) =\sqrt{(m_B)/(m_A) }`

Where;

`v_A` = The rate of diffusion of gas A

`v_B` = The rate of diffusion of gas B

`m_A` = The molar mass of the gas A

`m_B` = The molar mass of the gas B

Therefore, for ethane and gas Q, measured under the same condition, we have;

`(v_(ethane))/(v_Q) =\sqrt{(m_Q)/(m_(ethane)) }`

`(1 \ cm^3/s)/((100)/(121) \ cm^3/s) =\sqrt{(m_Q)/(30 \ g/mol) }`

`m_Q = \left ({(121)/(100) } \right) ^2 * 30 \ g/mol = 43.923 \ g/mol`

The molar mass of gas Q,
`m_Q` = 43.923 g/mol.