Question

At standard temperature and pressure, the volume of 7.1 kg of $C{l}_2$ will be:

• A. 2240 L
• B. 2.24 L
• C. 224 L
• D. 4.48 L

Answer 1

Answer: (A)

2240 L

Gram molecular weight of $C{l}_2$ (one mole)= 35.5 $\times$ 2 = 71 g.
71 g of $C{l}_2$ at S.T.P occupies $22.4$ litres.
$\therefore$ 7.1 kg, i.e., 7100 g of $C{l}_2$ would at STP occupy = $\frac{22.4\times 7100}{71}=2240L$