Question

At s.t.p. the density of CCL4 vapour in g/l will be what ?

Answer 1

Thanks for asking the question.
At same pressure and temperature, volume of one mole of a gas= 22.4l
1 ole contains 154grams of CCl4
As we know,
Density=mass\volume
d=154\22.4
d=6.875g\l
Hope it helps

Answer 2

At s.t.p. the density of $CCL_4$ vapour in g/l will be 6.875 g/l

Explanation:

Atomic masses of carbon and chlorine is 12 amu and 35.5 amu respectively.

So, the molecular mass of $CCL_4$ will be

$=12+4(35.5)$

= 154 u.

At Standard Temperature and Pressure, volume of the compound will be

$=n T \frac{R}{p}$

$=1 \times 273 \times \frac{0.0821}{1}$

= 22.4

Since, density $=\frac{\text { mass }}{\text { volume }}$

So, the density of $CCL_4$

$=\frac{154}{22.4}$

= 6.875  

Hence, the answer is 6.875 g/l.