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Question

Answers

(A) 0.48M

(B) 0.96M

(C) 0.24M

(D) 0.024M

Answer
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-The reactions involved here are:

$2C{H_3}COOH + CaOC{l_2} \to {(C{H_3}COO)_2}Ca + {H_2}O + C{l_2}$

$C{l_2} + 2KI \to 2KCl + {I_2}$

$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$

-Here we will be using the law of chemical equivalence.

According to this law whenever two substances react, the equivalents of one will be equal to the equivalents of the other and the equivalents of any product will be equal to the equivalents of the reactant.

For the reactions written above this law can be written as:

eq. of ${I_2}$ = eq. of $C{l_2}$ = eq. of ${S_2}{O_3}^{ - 2}$ = eq. of bleach ($CaOC{l_2}$) = eq. of $C{H_3}COOH$ (1)

-The question has given us the N and V of ${S_2}{O_3}^{ - 2}$ to be 0.25 N and 48 ml.

So, first we will calculate the equivalents of ${S_2}{O_3}^{ - 2}$:

eq. = N × V (ml) (2)

= 0.25 × 48

= 12 m.eq.

-Now from (2) we can say that: eq. of ${S_2}{O_3}^{ - 2}$ = eq. of $CaOC{l_2}$

So, the eq. of $CaOC{l_2}$ = 12 m.eq.

We also know that: (eq. = N × V) and (N = M × n-factor)

So, from this we can say that: eq. = M × n-factor × V (3)

For $CaOC{l_2}$ n-factor = 2

Hence we will now calculate the molar concentration of $CaOC{l_2}$ from equation (3):

eq. = M × n-factor × V

12 = M × 2 × 25

$M = \dfrac{{12}}{{2 \times 25}}$

= 0.24 M

Also, the household bleach has its properties due to the presence of chlorine which is a powerful oxidizer. They have broad spectrum bactericidal properties, which makes it highly useful for disinfecting, sterilizing and sanitizing purposes.