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 O2 at each temperature (assume the solution density is the same as water): Temperature (C∘) solubility of O2(mg/kgH2O) density of H2O (g/mL) 0.0 &nbs...

At 0∘Cmolarity of O2 is 4.53×10-4mol/L, at 20∘Cmolarity of O2 is 2.82×10-4mol/Land at molarity of O2 is 1.87×10-4mol/L.

Q13-121P - Chemistry: Molecular Nature Of Matter And Change

Q13-121P

Question

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

Step-by-Step Solution

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Answer

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 $1.87 \times 10^{- 4} m o l / L$ .

1Step 1: 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.

2Step 2: molarity of O 2 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

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

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

3Step 2: molarity of O 2 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

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

Q13-120CP

Q13-122CP

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