Solveeit Logo
Login

Question

Question: \( 5.0g \) of bleaching powder was suspended in water and volume made up to half a liter. \( 20ml \)...

5.0g5.0g of bleaching powder was suspended in water and volume made up to half a liter. 20ml20ml of this suspension when acidified with acetic acid and treated with excess potassium iodide solution liberated iodine ,which required 20ml20ml of a Deci-normal hypo solution for titration. Calculate the value of 10×x10 \times x , where x is the percentage of available chlorine in bleaching powder?

Explanation

Solution

Write down the chemical reaction and calculate the equivalent mass of chlorine and then relate them using N1V1=N2V2{N_1}{V_1} = {N_2}{V_2} after find out the mass of chlorine and then find the percentage.

Complete answer:
Let us first write down the chemical reaction for this question. In question it is given that bleaching powder is suspended in water to make up to half a liter of volume. For this step the reaction will be:
CaOCl2+H2O→Ca(OH)2+Cl2CaOC{l_2} + {H_2}O \to Ca{(OH)_2} + C{l_2}
Now this suspension was acidified and treated with excess of potassium iodide then the reaction will become:
Cl2+2KI→2KCl+I2C{l_2} + 2KI \to 2KCl + {I_2}
Now this product is titrated against a hypo solution then this will become:
I2+2Na2S2O3→Na2S4O6+NaI{I_2} + 2N{a_2}{S_2}{O_3} \to N{a_2}{S_4}{O_6} + NaI
Now using the concept of equivalence
1 mole CaOCl2=1 mole Cl2=1 mole I2=2 mole Na2S2O3{\text{1 mole CaOC}}{{\text{l}}_2} = 1{\text{ mole C}}{{\text{l}}_2} = 1{\text{ mole }}{{\text{I}}_2} = 2{\text{ mole N}}{{\text{a}}_2}{S_2}{O_3}
Now we will find the equivalent mass of chlorine
Cl2=712=35.5C{l_2} = \dfrac{{71}}{2} = 35.5
And N1V1(Cl2 in bleaching powder) = N2V2(hyposolution){N_1}{V_1}(C{l_2}{\text{ in bleaching powder) = }}{{\text{N}}_2}{V_2}(hypo solution) , now solving it we will get:
N1×20=0.1×20{N_1} \times 20 = 0.1 \times 20
N1=0.1N{N_1} = 0.1N
Now the normality of CaOCl2CaOC{l_2} solution will be 0.1N0.1N
Now we will find out the amount of chlorine in the pure sample.
It will be 0.1×35.5gL−1=3.55gL−10.1 \times 35.5g{L^{ - 1}} = 3.55g{L^{ - 1}}
The amount of impure sample in chlorine will be H
5.0g0.5L=10gL−1\dfrac{{5.0g}}{{0.5L}} = 10g{L^{ - 1}}
So the pure percentage of pure chlorine (x) available in the bleaching powder is :
3.5510×100=35.5%\dfrac{{3.55}}{{10}} \times 100 = 35.5\% this is our value of x
But in question it is asked to find the value of 10×x10 \times x so its value will be 10×35.5=35510 \times 35.5 = 355 which our answer is.

Note:
While using the concept of equivalence keep in mind that the equations you are using should be balanced for easiness and all the units should be in a system like if you are taking volume in liter then everywhere it should be in liter.