Q.

3.65 gm equimolar mixture of NaOH and Na2CO3 is titrated against 0.1 M HCI using phenolphthalein as an indicator, V1mL of acid was required to reach end point. In another experiment 3.65 gm of same mixture is titrated against 0.2 M HCl using methyl orange as an indicator. V2 mL  of acid was required to reach end point. V1 + V2 is:

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a

500 mL

b

875 mL

c

1000 mL

d

760 mL

answer is A.

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Detailed Solution

NaOH   Na2CO3

a molar   a molar

As per the first titration,

2a=0.1×V1.   V1=0.50.............. (i)

In the second experiment,

a+2a=0.2×V2

3.65=a×40+a×106

a=0.025

Putting the value of a in (1) we get,

 And V2=0.375

V1+V2=0.50+0.375=0.875ml

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