Question

Two identical balloons are inflated to a volume of 1.00 L with a particular gas. After 12 hours, the volume of one balloon has decreased by 0.200 L. In the same time, the volume of the other balloon has decreased by 0.0603 L. If the lighter of the two gases was helium, what is the molar mass of the heavier gas

Answer 1

43.7

Step-by-step explanation:

Volume of gas = 1.00L

Time = 12 hours

Decrease in 1 balloon = 0.200L

Decrease in second balloon = 0.0603L

R1 = rate at which helium diffuses= 0.20/12= 0.0166

R2 = rate at which heavier gas diffuses = 0.0693/12 = 0.00502

Using grahms law

R1²/R2² = m/4

= 0.0166²/0.00502² = m/4

= 10.935 = m/4

When we cross multiply

m is the molar mass of heavier gas = 43.73