Chapter 9, Problem 3QAP

The vapor pressure of I₂(s) at 30°C is 0.466 mm Hg.

(a) How many milligrams of iodine will sublime into an evacuated 750.0-mL flask?

(b) If 3.00 mg of I₂ are used, what will the final pressure in the flask be?

(c) If 7.85 mg of I₂ are used, what will the final pressure in the flask be?

Interpretation Introduction

Interpretation:

The mass of iodine (in mg) sublimed under the given conditions is to be calculated.

Concept Introduction :

The ideal gas equation is the expression that relates different measurable properties of a gas. The expression is given as

  $PV=nRT$

where,

P = Pressure of the gas

V = Volume of the container in which the gas is occupied

R = Ideal gas constant = 0.0821 L atm/ mol K

n = number of moles of the gas = Mass of gas / Molecular mass of gas

T = Absolute temperature of the gas i.e., temperature on Kelvin scale

Answer to Problem 3QAP

The mass of iodine sublimed under the given conditions = 2.41 mg.

Explanation of Solution

Given:

P = 0.466 mm Hg = (0.466/760) atm

V = 750 mL = 750×10-3 L

T = 30 ⁰C = (30+273) K = 303 K

When iodine sublimes, it forms vapors which are in equilibrium with the solid form. The pressure of the vapors in this state is vapor pressure. Thus, for gaseous camphor formed.

The number of moles of iodine present as vapors can be calculated by using the ideal gas equation as

  $\begin{array}{l}PV=nRT\\ \frac{0.466}{760}\text{ atm }\times \text{ 750}\times {\text{10}}^{-3}\text{ L= n (0}\text{.0821 L atm/mol K) (303 K)}\\ \text{n = 0}\text{.0185}\times {\text{10}}^{-3}\text{ mol}\end{array}$

The mass of iodine sublimed can thus be calculated as

  $\begin{array}{l}\text{number of moles = }\frac{\text{mass }}{\text{molecular mass}}\\ \text{Atomic mass of I = 127 g/mol}\\ \text{n = 0}\text{.0185}\times {\text{10}}^{\text{-3}}\text{ mol = }\frac{w}{\text{M}}=\frac{w}{[2\times 127]\text{ g/mol}}\\ \Rightarrow w=4.69\times {10}^{-3}\text{ g = 4}{\text{.69 mg (1 mg = 10}}^{\text{-3}}\text{ g)}\end{array}$

Therefore, 4.69 mg of iodine has sublimed under the given conditions.

Interpretation Introduction

Interpretation:

The pressure in the flask if 3 mg of iodine is taken needs be calculated.

Concept Introduction :

The ideal gas equation is the expression that relates different measurable properties of a

gas. The expression is given as:

  $PV=nRT$

where,

P = Pressure of the gas

V = Volume of the container in which the gas is occupied

R = Ideal gas constant = 0.0821 L atm/ mol K

n = number of moles of the gas ( $Numberofmoles=\frac{Givenmass}{Molecularmass}$ )

T = Absolute temperature of the gas i.e., temperature on Kelvin scale

Answer to Problem 3QAP

The pressure in the flask is 0.29 mm Hg.

Explanation of Solution

Given:

V = 750 mL = 750×10-3 L

Mass = m = 3 mg = 3×10-3 g

T = 30 ⁰C = (30+273) = 303 K

The pressure in the flask can be calculated using the ideal gas equation.

  $\begin{array}{l}PV=nRT\\ P\times \text{ 750}\times {\text{10}}^{-3}\text{ L= }\frac{3\times {10}^{-3}\text{ g}}{\text{[2}\times \text{127}]\text{ g/mol}}\text{ (0}\text{.0821 L atm/mol K) (303 K)}\\ \Rightarrow \text{P = 3}\text{.9}\times {\text{10}}^{-4}\text{ atm}\\ \Rightarrow \text{ P = 3}\text{.9}\times {\text{10}}^{-4}\times 760\text{ mm Hg = 0}\text{.29 mm Hg}\end{array}$

Thus, the pressure in the flask is 0.29 mm Hg.

Interpretation Introduction

Interpretation:

The pressure in the flask if 7.85 mg of iodine is taken needs to be calculated.

Concept Introduction :

The ideal gas equation is the expression that relates different measurable properties of a

gas. The expression is given as

  $PV=nRT$

where,

P = Pressure of the gas

V = Volume of the container in which the gas is occupied

R = Ideal gas constant = 0.0821 L atm/ mol K

n = number of moles of the gas ( $Numberofmoles=\frac{Givenmass}{Molecularmass}$ )

T = Absolute temperature of the gas i.e., temperature on Kelvin scale

Answer to Problem 3QAP

The pressure in the flask is 0.77 mm Hg.

Explanation of Solution

Given:

V = 750 mL = 750×10-3 L

Mass = m = 7.85 mg = 7.85×10-3 g

T = 30 ⁰C = (30+273) = 303 K

The pressure in the flask can be calculated using the ideal gas equation.

  $\begin{array}{l}PV=nRT\\ P\times \text{ 750}\times {\text{10}}^{-3}\text{ L= }\frac{7.85\times {10}^{-3}\text{ g}}{\text{[2}\times \text{127}]\text{ g/mol}}\text{ (0}\text{.0821 L atm/mol K) (303 K)}\\ \Rightarrow \text{P = 1}\text{.02}\times {\text{10}}^{-3}\text{ atm}\\ \Rightarrow \text{ P = 1}\text{.02}\times {\text{10}}^{-3}\times 760\text{ mm Hg = 0}\text{.77 mm Hg}\end{array}$

Thus, the pressure in the flask is 0.77 mm Hg.