Chapter 2: Q13-121P (page 40)

Thermal pollution from industrial wastewater causes the temperature of river or lake water to increase, which can affect fish survival as the concentration of dissolved O₂. decreases. Use the following data to find the molarity of O₂at each temperature (assume the solution density is the same as water):
Temperature ( ${}^{\circ}C$ ) solubility of O₂(mg/kgH₂O) density of H₂O (g/mL)
0.0 14.5 0.99987
20.0 9.07 0.99823
40.0 6.04 0.99224

Short Answer

Expert verified

At $0^{\circ} C$ molarity of O₂is $4.53 \times 10^{- 4} m o l / L$ , at $20^{\circ} C$ molarity of O₂is $2.82 \times 10^{- 4} m o l / L$ and at molarity of O₂is $1.87 \times 10^{- 4} m o l / L$ .

Step by step solution

Definitions

Molarity is defined as the number of moles per litre solution.

Solubility is defined as the maximum amount of substance that will dissolve in a given amount of solvent at a specific temperature.

The density of a substance is its mass in per unit volume.

molarity of O2 at 0∘C

It is given that the solubility and density of O₂change with change in temperature.

Given solubility = $\frac{(14.5 m g O_{2})}{(1 k g H_{2} O)} = \frac{(14.5 m g O_{2})}{(10^{6} m g H_{2} O)}$

One mole oxygen contains 32 grams of oxygen. Hence,

\frac{(1 m o l O_{2})}{(32 g O_{2})}

Given density of water = $\frac{(0.99987 g)}{m L} = \frac{0.99987 g}{(10^{- 3} L)}$

By using all the above factors molarity of oxygen can be calculated as follows

role="math" localid="1663392619761" $m o l a r i t y o f O_{2} = \frac{(14.5 m g O_{2})}{(10^{6} m g H_{2} O)} \times \frac{(1 m o l O_{2})}{(32 g O_{2})} \times \frac{0.99987 g}{(10^{- 3} L)}$

$= 4.53 \times 10^{- 4} m o l / L$

molarity of O2 at 40∘C

At 40⁰C given solubility = $\frac{(6.04 m g O_{2})}{(1 k g H_{2} O)} = \frac{(6.04 m g O_{2})}{(10^{6} m g H_{2} O)}$

Density of water = $\frac{0.99224 g}{m L} = \frac{0.99224 g}{10^{- 3} L}$

By using all the above factors molarity of oxygen can be calculated as follows

$m o l a r i t y o f O_{2} = \frac{6.40 m g O_{2}}{10^{6} m g H_{2} O} \times \frac{1 m o l O_{2}}{32 g O_{2}} \times \frac{0.99224 g}{10^{- 3} L} = 1.87 \times 10^{- 4} m o l / L$