What time will be required for a sample of ethane `(C_2 H_6)` to effuse through an orifice if a similar sample of butane `(C_4 H_(10)`) required 2 minutes 23 seconds ?

Explanation

To solve the problem of determining the time required for a sample of ethane `(C_2H_6)` to effuse through an orifice, given that a similar sample of butane `(C_4H_{10})` took 2 minutes and 23 seconds, we will use Graham's law of effusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be expressed as:
\[ \frac{\text{Rate of effusion of gas A}}{\text{Rate of effusion of gas B}} = \sqrt{\frac{M_B}{M_A}} \]
where \( M_A \) and \( M_B \) are the molar masses of gases A and B, respectively.
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