Respuesta :
Answer:
The molar mass of gas Q is 43.923 g/mol
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, [tex]v_{ethane}[/tex], is given as follows;
[tex]v_{ethane}[/tex] = (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, [tex]v_Q[/tex] = 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;
[tex]\dfrac{v_A}{v_B} =\sqrt{\dfrac{m_B}{m_A} }[/tex]
Where;
[tex]v_A[/tex] = The rate of diffusion of gas A
[tex]v_B[/tex] = The rate of diffusion of gas B
[tex]m_A[/tex] = The molar mass of the gas A
[tex]m_B[/tex] = The molar mass of the gas B
Therefore, for ethane and gas Q, measured under the same condition, we have;
[tex]\dfrac{v_{ethane}}{v_Q} =\sqrt{\dfrac{m_Q}{m_{ethane}} }[/tex]
[tex]\dfrac{1 \ cm^3/s}{\dfrac{100}{121} \ cm^3/s} =\sqrt{\dfrac{m_Q}{30 \ g/mol} }[/tex]
[tex]m_Q = \left ({\dfrac{121}{100} } \right) ^2 \times 30 \ g/mol = 43.923 \ g/mol[/tex]
The molar mass of gas Q, [tex]m_Q[/tex] = 43.923 g/mol.
