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Question

Bromine is prepared commercially by the reaction:
Br(aq)+Cl2(aq)Cl(aq)+Br2(aq)
suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.050 M solution of Cl2 is needed to react completely with the Br ?

A
50 mL
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B
10 mL
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C
25 mL
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D
30 mL
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Solution

The correct option is D 30 mL
50 mL of 0.060 M NaBr contain NaBr
=0.0601000×50=0.003 mol
2 mol of Br react with 1 mol of Cl2 because the balanced equation is 2Br+Cl22Cl+Br2
0.003 mol Br will react with =12×0.003
= 0.0015 mol Cl2
Given, 0.05 mol of Cl2 solution is present in 1000 mL of Cl2 solution
volume of Cl2=10000.05×0.0015=30 mL

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