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Question: The density of \( CC{l_4} \) vapor at \( 0^\circ C \) and \( 1atm \) in \( g/L \) is : (A) \( 11....

The density of CCl4CC{l_4} vapor at 0∘C0^\circ C and 1atm1atm in g/Lg/L is :
(A) 11.211.2
(B) 7777
(C) 6.876.87
(D) Can’t be found.

Explanation

Solution

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=nRTPV = 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=mMn = \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:
PV=mMRT PM=mVRT d=mV ∴PM=dRT \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 CCl4CC{l_4}
The conditions given in the question are the standard temperature and pressure so here
P=1atm T=273K \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 CCl4CC{l_4} =12+4×35.4=154.6≈154gm= 12 + 4 \times 35.4 = 154.6 \approx 154gm (C=12,Cl=35.4)(C = 12,Cl = 35.4)
So M=154gmM = 154gm
Thus substituting the values, we get:
1atm×154g/mol=d×0.0821Latmmol−1K−1×273K d=6.87g/L \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=MVD = \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&VM\& V in the above formula we get :
d=15422.4 d=6.87g/L \begin{gathered} d = \dfrac{{154}}{{22.4}} \\\ d = 6.87g/L \\\ \end{gathered}