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Question: Sea water contains around \( 65{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }}{L^{ - 1}} \) of br...

Sea water contains around 65 × 10−3g L−165{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }}{L^{ - 1}} of bromide ions. If all the bromide ions are converted into Br2B{r_2} , then how much sea water is needed to produce 1 kg1{\text{ }}kg of bromide Br2B{r_2} ?

Explanation

Solution

Here we are given the density of bromine in sea water. Thus it depicts the amount of bromine ion present in one litre of water. We will write the balanced conversion reaction of bromide ions into Br2B{r_2} and by using the concept of conservation of mass we can find the amount of sea water which will contain 1000 g1000{\text{ }}g or 1 kg1{\text{ }}kg of bromine.

Complete answer:
From the question we can say that the amount of bromide ions present in the water is 65 × 10−3g L−165{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }}{L^{ - 1}} , which means that 65 × 10−3g65{\text{ }} \times {\text{ }}{10^{ - 3}}g of bromide ions are present in one litre of water. It can be depicted as:
⇒65 × 10−3g L−1\Rightarrow 65{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }}{L^{ - 1}}
⇒65 × 10−3gL\Rightarrow 65{\text{ }} \times {\text{ }}{10^{ - 3}}\dfrac{g}{L}
Now we will write the balanced chemical equation for the conversion of bromide ions as:
Br− → 12Br2 + eB{r^ - }{\text{ }} \to {\text{ }}\dfrac{1}{2}B{r_2}{\text{ }} + {\text{ }}e
Now from the above reaction we can say that the number of bromide ions which are formed is half of the bromine. Thus we can write as:
⇒65 × 10−3g \Rightarrow 65{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }} of bromide ions (Br−)\left( {B{r^ - }} \right) will be equal to half of bromine.
⇒65 × 10−3g \Rightarrow 65{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }} of Br− = 12 × 65 × 10−3gB{r^ - }{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }} \times {\text{ }}65{\text{ }} \times {\text{ }}{10^{ - 3}}g of Br2B{r_2}
⇒65 × 10−3g \Rightarrow 65{\text{ }} \times {\text{ }}{10^{ - 3}}g{\text{ }} of Br− = 32.5 × 10−3gB{r^ - }{\text{ }} = {\text{ 32}}{\text{.5 }} \times {\text{ }}{10^{ - 3}}g of Br2B{r_2}
Therefore we can say that,
⇒1000 g\Rightarrow 1000{\text{ }}g or 1 kg1{\text{ }}kg of Br2B{r_2} will be present in = 1000 g 32.5 × 10−3g= {\text{ }}\dfrac{{1000{\text{ }}g}}{{{\text{ 32}}{\text{.5 }} \times {\text{ }}{{10}^{ - 3}}g}}
⇒1000 g\Rightarrow 1000{\text{ }}g or 1 kg1{\text{ }}kg of Br2B{r_2} will be present in = 30769= {\text{ 30769}}
Thus we required 30769 L{\text{30769 L}} of water to produce 1  1{\text{ }}kg of bromide Br2B{r_2} .

Note:
The above method which we used is known as the unitary method to find the amount of substance. The conversion reaction of bromide ion to bromine must be a balanced chemical reaction. Also bromide ion is formed when a bromine atom accepts one electron and becomes inert in the atmosphere by achieving inert gas configuration. Density of substance is the ratio of the amount of substance present in one litre of solution.