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Question:

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be:

36u

64u

9u

27u

Solution:

A certain gas takes three times as long to effuse out as helium. The rate of effusion is inversely proportional to the square root of the molecular mass. This is Graham's law of effusion. Mathematically, it can be expressed as:

Rate₁ / Rate₂ = √(M₂ / M₁)

Where:

  • Rate₁ is the rate of effusion of gas 1
  • Rate₂ is the rate of effusion of gas 2
  • M₁ is the molar mass of gas 1
  • M₂ is the molar mass of gas 2

Let's denote the rate of effusion of helium as Rate(He) and the rate of effusion of the unknown gas as Rate(X). We are given that the unknown gas takes three times as long to effuse, meaning its rate is one-third that of helium:

Rate(X) = (1/3)Rate(He)

The molar mass of helium (He) is 4 u. Let's denote the molar mass of the unknown gas as M(X). Applying Graham's law:

Rate(He) / Rate(X) = √(M(X) / 4u)

Substituting Rate(X) = (1/3)Rate(He):

Rate(He) / [(1/3)Rate(He)] = √(M(X) / 4u)

3 = √(M(X) / 4u)

Squaring both sides:

9 = M(X) / 4u

M(X) = 9 * 4u = 36u

Therefore, the molecular mass of the unknown gas is 36u.