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In carius method of estimation of bromine, it was observed that the percentage of bromine present in an organic compound is 50%. Ratio between AgBr(Mol.wt = 188) precipitated to mass of organic compound taken is

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1.175

11.75

0.1175

117.5

answer is A.

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Detailed Solution

$% Br = \frac{80}{188} \times \frac{mass of AgBr precipitated}{mass of organic compound taken} \times 100$

$\frac{50 \times 188}{80 \times 100} = \frac{mass of AgBr precipitated}{mass of organic compound taken}$

$\frac{188}{160} = \frac{mass of AgBr precipitated}{mass of organic compound taken}$

=1.175

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