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Question

50 mL of a mixture of NaOH and Na2CO3 titrated with
N10 HCl using phenolpthalien indicator required 50ml of HCl to decolour isephenolpthalien. At this stage methyl orange was added and addition of acid was continued. The second end point was reached after further addition of 10ml of N10. The amount of NaOH in the solution is:


A
3.2g
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B
0.16g
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C
0.08g
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D
0.4g
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Solution

The correct option is B 0.16g
1st titration
Meq. Of NaOH + 12 meq of Na2CO3 = meq of HCL = 50×110=5
2nd titration
12 meq of Na2CO3 = meq. of HCL = 10×110=1
M eq of NaOH = 5 - 1 = 4
Lot of NaOH = 4×40100=0.16gr

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