At S.T.P. the density of $ ext{CCl}_4$ vapour in $ ext{g/L}$ will be nearest to:

Question

At S.T.P. the density of $	ext{CCl}_4$ vapour in $	ext{g/L}$ will be nearest to:

Accepted Answer

At Standard Temperature and Pressure (STP), $1	ext{ mole}$ of an ideal gas occupies a volume of $22.4	ext{ L}$.
Molar mass of $	ext{CCl}_4 = 12 + 4(35.5) = 154	ext{ g/mol}$.
Density is the mass per unit volume. Therefore, the density of $	ext{CCl}_4$ vapour at STP is:
$	ext{Density} = \frac{	ext{Molar Mass}}{	ext{Molar Volume}} = \frac{154	ext{ g/mol}}{22.4	ext{ L/mol}} = 6.875	ext{ g/L}$.
The calculated density is $6.875	ext{ g/L}$, which is nearest to the option $6.84$.

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