An optically active substance decomposes into optically active substances ‘B’ and ‘C’ as follows
$A \overset{k = 0.001 \min u t e^{- 1}}{\to} 2 B + C$
The specific rotations of A,B and C are $+ 40^{0}, + 10^{0} a n d - 30^{0}$ per mole, respectively. If initially A and C were present in 4:3 mole ratio, the time (in min), after which the sample becomes optically inactive is $(\ln 2 = 0.7, \ln 5 = 1.6, \ln 7 = 2.0, \ln 13 = 2.5)$

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answer is 500.

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Detailed Solution

ANS-500
SOL-
$A \to 2 B + C t = 0 4 a 0 3 a t = t 4 a - x 2 x 3 a + x$
From the question
$(4 a - x) (40^{°}) + 2 x (10^{°}) + (3 a + x) (- 30^{°}) = 0^{°}$
$∴ x = \frac{7}{5} a$
Now,
$t = \frac{1}{K} \times \ln \frac{4 a}{4 a - \frac{7}{5} a} = \frac{1}{0.001}, \ln \frac{20}{13} = 500 \min$