In Carius method, $0.099 g m$ organic compound gave $0.287 g m A g C l$ . The percentage of chlorine in the compound will be:

28.6

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71.7

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35.4

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64.2

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Solution

The correct option is B $71.7$
We know, $0.287 gm A g C l = \frac{0.287}{143.5} m o l A g C l = 0.002 m o l A g C l$

Again, $0.002 mol chlorine = 0.071 gm chlorine$

So the percentage of chlorine in the organic compound will be,
$= \frac{0.071}{0.099} \times 100 \approx 71.7 %$

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In Carius method, 0.099 gm organic compound gave 0.287 gm AgCl. The percentage of chlorine in the compound will be:

The correct option is B 71.7We know, 0.287 gm AgCl=0.287143.5 mol AgCl=0.002 mol AgCl Again, 0.002 mol chlorine = 0.071 gm chlorine So the percentage of chlorine in the organic compound will be, =0.0710.099×100≈71.7%

In Carius method, $0.099 g m$ organic compound gave $0.287 g m A g C l$ . The percentage of chlorine in the compound will be:

The correct option is B $71.7$
We know, $0.287 gm A g C l = \frac{0.287}{143.5} m o l A g C l = 0.002 m o l A g C l$

Again, $0.002 mol chlorine = 0.071 gm chlorine$

So the percentage of chlorine in the organic compound will be,
$= \frac{0.071}{0.099} \times 100 \approx 71.7 %$