Chapter 19, Problem 3STP

Interpretation Introduction

Interpretation:

Molarity of four gases is given. The container containing $110g$ of gas is to be find out.

Concept introduction:

The question is based on the concept molarity.

Molarity is ratio of number of moles of solute to the volume of solution in liters.

$Molarity=\frac{numberofmolesofsolute}{volumeofsolutioninliters}$

Molarity is expressed in $mol/L$

Further, $\begin{array}{l}Numberofmoles=\raisebox{1ex}{$givenmass$}\!\left/ \!\raisebox{-1ex}{$molarmass$}\right.\\ \therefore Molarity=\frac{givenmassofsolute}{molarmassofsolute\times volumeofsolutioninliters}\\ Molarity=\frac{m}{M\times V}\\ where\\ misgivenmassofsolute\\ Mismolarmassofsolute\\ Visvolumeofsolutioninliters\\ Therefore,\\ m=molarity\times M\times V\end{array}$

Answer to Problem 3STP

D

Explanation of Solution

For Xe:

$\begin{array}{l}Molarity=0.50mol/L\\ M=131\\ V=5L\end{array}$

Therefore, by formula:

$\begin{array}{l}m=molarity\times M\times V\\ m=0.5\times 131\times 5\\ =327.5g\end{array}$

For He:

$\begin{array}{l}Molarity=0.50mol/L\\ M=4\\ V=5L\end{array}$

Therefore, by formula:

$\begin{array}{l}m=molarity\times M\times V\\ m=0.5\times 4\times 5\\ =10g\end{array}$

For $N_2$:

$\begin{array}{l}Molarity=0.50mol/L\\ M=28\\ V=5L\end{array}$

Therefore, by formula:

$\begin{array}{l}m=molarity\times M\times V\\ m=0.5\times 28\times 5\\ =70g\end{array}$

For $C{O}_2$:

$\begin{array}{l}Molarity=0.50mol/L\\ M=44\\ V=5L\end{array}$

Therefore, by formula:

$\begin{array}{l}m=molarity\times M\times V\\ m=0.5\times 44\times 5\\ =110g\end{array}$

Thus $C{O}_2$ gas is 110 g.

Conclusion

Mathematical calculations show that gas D has 110 g mass. Therefore, option D is correct and others are incorrect.