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Q.

0.05 mole of CaOCl2 reacts with excess of and liberated is I2 titrated with V mL of 0.1 mL hypo solution then value of V is

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a

500 mL

b

1000 mL

c

2000 mL

d

250 mL

answer is B.

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Detailed Solution

Number equivalent of hypo = Number equivalent of   CaOCl2M1V1=M2V2⇒0.05×2×100=0.1×V2          (∴nf of CaOCl2=2)∴V2=1000 mL
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