Question

At STP, the density of $CCl_4$ vapour in g/L will be nearest to

• A. 6.87
• B. 3.42
• C. 10.26
• D. 4.57

Answer 1

Answer: (A)

6.87

Answer 2

1 mole $CC1_4 \, vapours = 12 + 4 \times 35.5 = 154 g$
At STP, volume of 1 mole of a gas = 22.4 L
Thus, $\, \, \, \, \, \, \, \, \, \, \, \, 154 \, g \, = 22.4L$
$\therefore$ Density of $CCl_4$ vapours $= \frac {154}{22.4}gL^{-1}$
$= 6.87 g L^{-1}$