Question

3.7 g of a gas at 25°C occupied the same volume as 0.184g of hydrogen at 17°C and at the same pressure. What is the molecular weight of the gas?

Answer 1

✔ W1 = Weight of unknown gas.

✔ W2 = Weight of hydrogen gas.

✔ M1 = molar mass of unknown gas.

✔ M2 = molar mass of hydrogen gas.

✔ T1 = temperature of unknown gas.

✔ T2 = temperature of hydrogen gas.

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$\bold{UNKNOWN\:\:GAS\:\: :}$

(A) Weight (W1) = 3.7 g

(B) Temperature attained (T1) = 298 K

So, Moles of this gas = W1/M1 = 3.7/M1

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$\bold{HYDROGEN\:\:GAS\:\: :}$

(A) Weight (W2) = 0.184 g

(B) Temperature attained = 290 K

Moles of hydrogen = W2/M2

So, moles = 0.184/2 = 0.092

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Applying ideal gas equation, we get :

HYDROGEN : P1V1 = n1RT1

UNKNOWN GAS : P2V2 = n2RT2

Thus,

P1V1/P2V2 = n1RT1/n2RT2

By Boyle's law, we know, P1V1 = P2V2 = constant (k).

Or, k/k = (0.092×298)/(3.7/M1 ×290)

Or, 1 = 27.416/(1073/M1)

Or, 1 = (27.416 × M1)/1073

Or, M1 = 1073/27.416

Or, M1 = 39.15