Answer to Question #145868 in General Chemistry for Erica

Question

Find the molar mass of an unknown gas which diffuses twice as fast as Cl2 gas.

Answer 1

Q145868

Find the molar mass of an unknown gas which diffuses twice as fast as Cl₂ gas.

Solution:

Graham’s Law: Rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molar mass.

For two gases A and B, their rate of diffusion and molar mass can be related by the formula

${{Rate_A}\over{Rate _B}} = \sqrt({ {M_B}\over{M_A} })$

where, Rate _A = rate of diffusion of gas A.

M_A = molar mass of gas A.

Rate _B = rate of diffusion of gas B

M_B = molar mass of gas B.

Suppose the unknown gas is ‘X’ and the molar mass of the unknown gas is M_X.

Molar mass of Cl₂ = 2 * atomic mass of Cl = 2 * 35.453 g/mol

= 70.906 g/mol

In Question, we are given that the diffusion rate of the unknown gas is twice as fast as Cl₂ gas.

Which means

Rate of diffusion of unknown gas = 2 * Rate of diffusion of Cl₂ .

Rate _X = 2 * Rate _Cl2 .

which can also be written as

${{Rate_X}\over{Rate _{Cl2}}} = 2;$

plug this and the molar mass of Cl₂ in Graham’s equation, we have  

${{Rate_X}\over{Rate _{Cl2}}} = \sqrt({ {M_{Cl2}}\over{M_X} }) ;$

$2 = \sqrt({ {70.906g/mol}\over{M_X} }) ;$

squaring both sides of the equation, we have  

$4 ={ {70.906g/mol}\over{M_X} } ;$

which can also be written as  

$M_X ={ {70.906g/mol}\over{4} } ;$

$M_X = 17.73g/mol$

Hence the molar mass of the unknown gas is 17.73g/mol.