60cm3 of hydrogen diffused through a porous membrane in 10minutes.The same volume of gas G diffused through the same membrane in 37.4 minutes .Determine the relative molecular mass of G.(H=1)
rateG/rateH2 = sqrt (molmassH2/molemassG)
first, relative molmass H2=1
so molmassG=(rateH2/rateG)^2
but rates are inversley proportion to times of diffusion, so
relative molmassG=(3.74(^2=13.98 relative to H=1
actual molmass = twice that
Exactly
To determine the relative molecular mass of gas G, we can use Graham's law of diffusion.
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, we can represent Graham's law as:
Rate1/Rate2 = √(Molar Mass2/Molar Mass1)
In this case, hydrogen (H2) is gas 1 and gas G is gas 2.
Given that 60 cm3 of hydrogen diffused through the membrane in 10 minutes, and the same volume of gas G diffused through the same membrane in 37.4 minutes, we can set up the equation:
(60 cm3/10 minutes) / (60 cm3/37.4 minutes) = √(Molar Mass of G/Molar Mass of Hydrogen)
Simplifying this equation:
(60/10) / (60/37.4) = √(Molar Mass of G/1)
6/1.608 = √(Molar Mass of G/1)
Taking the square of both sides:
(6/1.608)^2 = (Molar Mass of G/1)^2
36/2.587264 = (Molar Mass of G/1)^2
13.9312 = (Molar Mass of G/1)^2
Taking the square root of both sides:
√13.9312 = (Molar Mass of G/1)
3.7343 = (Molar Mass of G/1)
Therefore, the relative molecular mass of gas G is approximately 3.7343 (rounded to four decimal places).
To determine the relative molecular mass of gas G, we will use Graham's law of diffusion. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
First, let's determine the rate of diffusion for each gas. The rate of diffusion is calculated by dividing the volume of gas diffused by the time taken for diffusion.
For hydrogen (H2):
Rate of diffusion = Volume of hydrogen diffused / Time taken for diffusion
= 60 cm^3 / 10 min
= 6 cm^3/min
For gas G:
Rate of diffusion = Volume of gas G diffused / Time taken for diffusion
= 60 cm^3 / 37.4 min
= 1.604 cm^3/min
Next, we will use Graham's law to compare the rate of diffusion of hydrogen and gas G. According to Graham's law, the ratio of the rates of diffusion is equal to the square root of the ratio of the molar masses.
Let's assume the molar mass of G is M.
Therefore, we can write the ratio of rates of diffusion as:
6 cm^3/min (rate of hydrogen) / 1.604 cm^3/min (rate of G) = √(Molar mass of G / Molar mass of hydrogen)
Since the molar mass of hydrogen (H2) is 2 g/mol, we can substitute these values into the equation:
6 / 1.604 = √(M / 2)
Now, let's solve for M:
36 / 1.604^2 = M / 2
36 * 2 = M * 1.604^2
M * 1.604^2 = 72
M = 72 / 1.604^2
Using a calculator, M ≈ 27.990 g/mol (rounded to three decimal places)
Therefore, the relative molecular mass of gas G is approximately 27.990 g/mol.