Mole ConceptHard

Question

A quantity of 10 g of a mixture of C2H6 and C5H10 occupy 4480 ml at 1 atm and 273 K. The percentage of C2H6 by mass in the mixture is

Options

A.30%
B.70%
C.50%
D.60%

Solution

Average molar mass of the mixture $= \frac{22400}{4480} \times 10 = 50$

Now, $\frac{100}{50} = \frac{x}{30} + \frac{100 - x}{70}$

⇒ Mass per cent of C2H6, x = 30

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