Chapter 10, Problem 6STP

Interpretation Introduction

Interpretation: The true statements for a balloon filled with 1.00 mol nitrogen gas at STP need to be identified.

Concept Introduction: The equation for an ideal gas is as follows:

  $PV=nRT$

Here, P is pressure, V is volume, n is the number of moles, R is the Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Answer to Problem 6STP

  ${\text{ΔH}}_{\text{(reaction)}}\text{ = -127 kJ/mol}$

Explanation of Solution

Reason for correct option:

At STP, the volume of the balloon can be calculated as follows:

  $PV=nRT$

Or,

  $V=\frac{nRT}{P}$

The standard temperature and pressure values are $273.15\text{ K}$ and 1 atm respectively. Thus,

  $\begin{array}{c}V=\frac{\left(1\text{ mol}\right)\left(0.0823\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}{\left(1\text{ atm}\right)}\\ =22.4\text{ L}\end{array}$

Thus, the volume of the balloon is 22.4 L. Therefore, the statement I is true.

Now, the number of moles of nitrogen gas in the balloon is 1 mol and the molar mass of nitrogen gas is 28 g/mol; thus, the mass of nitrogen gas will be:

  $m=n\times M$

Putting the values,

  $\begin{array}{c}m=\left(1\text{ mol}\right)\left(28\text{ g/mol}\right)\\ =28\text{ g}\end{array}$

Thus, the mass of content in the balloon is 28 g. Therefore, statement II is not true.

According to Avogadro’s law, 1 mol of any substance contains $6.022\times {10}^{23}$ units thus, 1 mol of nitrogen gas contains $6.022\times {10}^{23}$ molecules of nitrogen gas. Therefore, statement III is true,

The true statements are statements I and III only. Therefore, option (C) is the correct option.

Conclusion

Reason for incorrect options:

Since only statement I is not true both statements I and III are true, thus, option (A) is the incorrect option.

Statement II is not true; thus, option (B) and option (D) are incorrect options.