Question

A 45 ml (Specific gravity 1.02) of chlorine water is treated with an excess of KI. The liberated iodine requires 26 ml of 0.15 N $N{a_2}{S_2}{O_3}$ (sodium thiosulphate or hypo) solution. What is the percentage of $C{l_2}$ (by mass) in the chlorine water? Chlorine water is a solution of free chlorine in water.
A. 0.388%
B. 0.301%
C. 0.307%
D. 3.02%

Answer 1

Hint: To find mass percentage of chlorine, we need mass of chlorine and total mass of solution. We can get Total mass of solution from formula:
\[density = \dfrac{{mass}}{{Volume}}\]
Mass of chlorine can be obtained by taking equivalent mass of Iodine equal to equivalent mass of chlorine. And equivalent mass of Iodine can be obtained from given data of sodium thiosulphate.

Complete step by step answer:
Firstly, we need to get the total mass of the solution using the formula of density.
\[density = \dfrac{{mass}}{{Volume}}\]
Now, we know specific gravity= density = 1.02 (given in question)
Volume of chlorine solution = 45 ml
Now substitute in above equation, we get:
\[1.02 = \dfrac{{mass}}{{45}}\]
Cross multiplying, we can get mass of solution, as:
\[\therefore {\text{mass of solution = }}45 \times 1.02 = 45.9gm\] ------ equation 1
Now, to get mass of chlorine we have to use the concept of equivalent mass.
\[Equivalent{\text{ }}mass{\text{ }}of\;N{a_2}{S_2}{O_3} = {\text{ }}Normality \times Volume\]
Normality = 0.15 N (given)
Volume of $N{a_2}{S_2}{O_3}$ = 26 ml
Substitute these values to get equivalent mass of $N{a_2}{S_2}{O_3}$
\[{\text{Equivalent mass of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 0.15 \times 26 \times {10^{ - 3}} = 3.9 \times {10^{ - 3}}\]
Equivalent mass of Iodine = Equivalent mass of sodium thiosulphate = \[3.9 \times {10^{ - 3}}\]
The reaction of chlorine water with KI can be written as:
\[C{l_2} + 2KI \to 2KCl + {I_2}\]
From the above reaction, we can say that the equivalent mass of Iodine = Equivalent mass of Chlorine.
\[\therefore {\text{Equivalent mass of Chlorine = 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}\]
We know formula of equivalent mass,
\[{\text{Equivalent mass of Chlorine = }}\dfrac{{{\text{mass of chlorine}}}}{{{\text{Atomic mass of chlorine}}}}\]
Now, we know, Atomic mass of chlorine=35.5, and substitute in above equation, we get:
\[3.9 \times {10^{ - 3}} = \dfrac{{{\text{mass of chlorine}}}}{{35.5}}\]
Cross multiply, to get mass of chlorine.
\[{\text{mass of chlorine = 35}}{\text{.5}} \times {\text{3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}\]
On simplification, we get
\[\therefore {\text{mass of chlorine = 0}}{\text{.138 gm}}\] ------- equation 2
Now, to get percentage of chlorine, the formula used is:
\[\% chlorine = \dfrac{{{\text{mass of chlorine}}}}{{{\text{mass of solution}}}} \times 100\]
Substitute equation 1 and equation 2, we get:
\[\% chlorine = \dfrac{{0.138}}{{45.9}} \times 100\]
On simplification, we get chlorine percentage as
\[\% chlorine = 0.301\% \]
Hence, the percentage of chlorine in chlorine water is 0.301%

Hence the correct option is B.

Note: We should take care of conversion of ml to litres while substituting volume with normality. The concept of equivalent mass is essential to be known, which does not even depend on stoichiometric coefficients. Specific gravity is same as density, and the unit of density is \[g{\text{ }}m{l^{ - 1}}\]