Question

The density of $ CC{l_4} $ vapor at $ 0^\circ C $ and $ 1atm $ in $ g/L $ is :
(A) $ 11.2 $
(B) $ 77 $
(C) $ 6.87 $
(D) Can’t be found.

Answer 1

Hint :Density can be stated as mass per unit volume of a particular substance. It is well known that all gases occupy the same volume on a per mole basis so the density of any specific gas depends upon its molar mass. Therefore, it can be said that the gas possessing a small molar mass tends to have a low density in comparison to a gas possessing a large molar mass.

Complete Step By Step Answer:
At STP (i.e. Standard Temperature and Pressure), we can use the ideal gas equation to calculate the density of gas in g/litre:
 $ PV = nRT $
At STP:
 $ \begin{array}{*{20}{l}}
  {P = 1{\text{ }}atm} \\
  {T = 273{\text{ }}K} \\
  {R = gas{\text{ }}constant = 0.0821{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}}
\end{array} $
Number of moles, ‘n’ can be calculated using the following formula:
 $ n = \dfrac{m}{M} $
m = given mass
M = molar mass
Density of a gas can be calculated as the ratio of given mass and volume.
Substituting this in the ideal gas equation, we get:
 $ \begin{gathered}
  PV = \dfrac{m}{M}RT \\
  PM = \dfrac{m}{V}RT \\
  d = \dfrac{m}{V} \\
  \therefore PM = dRT \\
\end{gathered} $
 Using this formula, we will calculate the density of $ CC{l_4} $
The conditions given in the question are the standard temperature and pressure so here
 $ \begin{gathered}
  P = 1atm \\
  T = 273K \\
\end{gathered} $
Here we have converted the temperature from degree to kelvin.
We will first calculate the molar mass of $ CC{l_4} $ $ = 12 + 4 \times 35.4 = 154.6 \approx 154gm $ $ (C = 12,Cl = 35.4) $
So $ M = 154gm $
Thus substituting the values, we get:
 $ \begin{gathered}
  1atm \times 154g/mol = d \times 0.0821Latmmo{l^{ - 1}}{K^{ - 1}} \times 273K \\
  d = 6.87g/L \\
\end{gathered} $
Thus, the correct answer is Option C.

Note :
Alternatively, density of a gas can be calculated as the ratio of molar mass and molar volume at STP. The following formula can be used to calculate the density of gas:
 $ D = \dfrac{M}{V} $
D = density of gas at STP
M = molar mass
V = molar volume of a gas at STP (22.4 L/mol)
In the present question, it will be as follows:
On putting the values of $ M\& V $ in the above formula we get :
 $ \begin{gathered}
  d = \dfrac{{154}}{{22.4}} \\
  d = 6.87g/L \\
\end{gathered} $