if 100cm3 of O2 diffused in 4 seconds and 50cm3 of gas Y diffuse in 3 seconds, calculate the relative molecular mass of gas Y?
Your answer is very wrong solve it again
The answer is 72.0
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To calculate the relative molecular mass of gas Y, we can use Graham's Law of Diffusion. According to Graham's Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
We are given that 100cm3 of O2 diffuses in 4 seconds and 50cm3 of gas Y diffuses in 3 seconds. Let's use these values to set up a ratio:
(rate of diffusion of O2) / (rate of diffusion of gas Y) = √(molar mass of gas Y) / √(molar mass of O2)
Substituting the given values:
(100 cm3 / 4 s) / (50 cm3 / 3 s) = √(molar mass of gas Y) / √(32 g/mol)
Simplifying:
(100 cm3 * 3 s) / (4 s * 50 cm3) = √(molar mass of gas Y) / √(32 g/mol)
(300 cm3/s) / (200 cm3/s) = √(molar mass of gas Y) / √(32 g/mol)
1.5 = √(molar mass of gas Y) / √(32 g/mol)
Squaring both sides:
(1.5)^2 = (√(molar mass of gas Y))^2 / (√(32 g/mol))^2
2.25 = (molar mass of gas Y) / 32
Rearranging the equation to solve for the molar mass of gas Y:
(molar mass of gas Y) = 2.25 * 32
(molar mass of gas Y) = 72 g/mol
Therefore, the relative molecular mass of gas Y is 72 g/mol.
T2/T=√M1√M2
T2=3 T1=4 M1=O2=32
M2=?
4/2=√M2/√32
=1.3