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Question: Tetrafluoroethylene, \({C_2}{F_4}\) , effuses through a barrier at a rate of \(8 \times {10^{ - 6}}m...

Tetrafluoroethylene, C2F4{C_2}{F_4} , effuses through a barrier at a rate of 8×106mol/hr8 \times {10^{ - 6}}mol/hr . An unknown gas (XX ) consisting only of boron and hydrogen; effuses at the rate of 10×106mol/hr10 \times {10^{ - 6}}mol/hr under the same conditions. The molecular mass of XX is [At. Weight of F=19amuF = 19amu ].

Explanation

Solution

Graham's law of effusion (also called Graham's law of diffusion) was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles or the molar mass of the gas.

Complete step by step answer:
Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate of the other (heavier gases diffuse more slowly). Mathematically, Graham’s law can be represented as:
R1MwR \propto \dfrac{1}{{\sqrt {{M_w}} }}
Where, R=R = rate of effusion of gas
Mw={M_w} = molecular weight of the gas
Thus, for tetrafluoroethylene, C2F4{C_2}{F_4}, and the unknown gas, the mathematical relation can be represented by:
RC2F4RX=MXMC2F4\dfrac{{{R_{{C_2}{F_4}}}}}{{{R_X}}} = \dfrac{{\sqrt {{M_X}} }}{{\sqrt {{M_{{C_2}{F_4}}}} }}…(i)
Where, RC2F4=8×106mol/hr{R_{{C_2}{F_4}}} = 8 \times {10^{ - 6}}mol/hr = rate of effusion of tetrafluoroethylene gas
RX=10×106mol/hr{R_X} = 10 \times {10^{ - 6}}mol/hr= rate of effusion of unknown gas
MC2F4=(2×12)+(4×19)=100amu{M_{{C_2}{F_4}}} = (2 \times 12) + (4 \times 19) = 100amu = molecular mass of tetrafluoroethylene gas
MX=?{M_X} = ? = molecular mass of unknown gas
Substituting the values in the equation (i), we have:
8×10610×106=MX100\Rightarrow \dfrac{{8 \times {{10}^{ - 6}}}}{{10 \times {{10}^{ - 6}}}} = \sqrt {\dfrac{{{M_X}}}{{100}}}
Thus, on solving, we have:
64100=MX100\Rightarrow \dfrac{{64}}{{100}} = \dfrac{{{M_X}}}{{100}}
Hence, MX=64amu{M_X} = 64amu .

Note:
A complete theoretical explanation of Graham's law was provided years later by the kinetic theory of gases. Graham's law provides a basis for separating isotopes by diffusion which is a method that came to play a crucial role in the development of the atomic bomb.