Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 29 s for 1.0 L of O2 gas to effuse.

Answer 1

Answer:

594.53 g/mol is the molar mass of the unknown gas.

Explanation:

The volume of the unknown gas  effusing out = V = 1.0 L

Time taken by 1 L of gas to effuse out = t = 125 s

Effusion rate of the unknown gas = $R=\frac{V}{t}=\frac{1.0 L}{125s}$

Volume of the nitrogen gas effusing out = V' = 1.0 L

Time taken by 1 L of oxygen gas to effuse out = t' = 29 s

Effusion rate of the oxygen gas = $r=\frac{V'}{t'}=\frac{1.0 L}{29 s}$

Molar mass of unknown gas = M

Mass of oxygen gas = 32 g/mol

Graham's law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{R}{r}=\sqrt{\frac{28 g/mol}{M}}$

$\frac{\frac{V}{125s}}{\frac{V'}{29s}}=\sqrt{\frac{32g/mol}{M}}$

$\frac{29}{125}=\sqrt{\frac{32g/mol}{M}}$

$M=\frac{32g/mol\times 125\times 125}{29\times 29}=594.53 g/mol$

594.53 g/mol is the molar mass of the unknown gas.

Answer 2

Answer:

The molar mass of the other gas is 594.53 g/mol

Explanation:

The complete question is:  A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 29 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

Step 1: Data given

It required 125 s for 1.0 L of the gas to effuse.

Under identical experimental conditions it required 29 s for 1.0 L of O2 gas to effuse.

Step 2: Calculate the molar mass

Graham's law of effusion says:

r1/r2 = √(M2/M1)

⇒ with r1 = effusion rate 1 = L/125s

⇒ with r2 = effusion rate 2 = L/29s

⇒ with M1 = molar mass of the gas = x g/mol

⇒ with M2 = molar mass of oxygen = 32 g/mol

We should be careful if we plug in the effusion rate. Because the fractions could flip if we don't give the effusion rates as numeral values.

(L/125s)/(L/29s) =  29/125 = √(32/M1)

29/125 =√(32/X)

(29/125)² = 32/X

0.053824 =32/X

X = 32/0.053824

X = 594.53 g/mol

The molar mass of the other gas is 594.53 g/mol