The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (1st order) were as follows :Time(min)030∞Rotation due to solution60°40°-30°Which of the following is correct? log 2=0.3, log 3=0.48, log 7=0.84, loge 10=2.3

Here,k×30=ln 60+3040+30n=5 ⇒k=9.2×10−3At, rt=0; 9.2×10−3×t=ln9030 ⇒t=120 min

The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (1st order) were as follows :Time(min)030∞Rotation due to solution60°40°-30°Which of the following is correct? log 2=0.3, log 3=0.48, log 7=0.84, loge 10=2.3

Here,k×30=ln 60+3040+30n=5 ⇒k=9.2×10−3At, rt=0; 9.2×10−3×t=ln9030 ⇒t=120 min

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The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (1st order) were as follows :
Time(min) 0 30 $\infty$
Rotation due to solution $60 °$ $40 °$ $- 30 °$
Which of the following is correct?
$[\log 2 = 0.3, \log 3 = 0.48, \log 7 = 0.84, \log_{e} 10 = 2.3]$

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The half life of reaction is 30 min.

The solution becomes optically inactive at 120 min.

Rotation after 40 sec. and after 80 sec will be in the opposite direction.

The time taken for $\frac{1}{4} th$ of sucrose to invert will 4 be 150 min.

answer is B.

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Detailed Solution

Here,

$k \times 30 = \ln {[\frac{60 + 30}{40 + 30}]}^{n = 5} \Rightarrow k = 9 .2 \times 10^{- 3}$

At, $r_{t} = 0;$

$9 .2 \times 10^{- 3} \times t = \ln (\frac{90}{30}) \Rightarrow t = 120 \min$

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Time(min)	0	30	$\infty$
Rotation due to solution	$60 °$	$40 °$	$- 30 °$

The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (1st order) were as follows :Time(min)030∞Rotation due to solution60°40°-30°Which of the following is correct? log 2=0.3, log 3=0.48, log 7=0.84, loge 10=2.3

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The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (1st order) were as follows :
Time(min) 0 30 $\infty$
Rotation due to solution $60 °$ $40 °$ $- 30 °$
Which of the following is correct?
$[\log 2 = 0.3, \log 3 = 0.48, \log 7 = 0.84, \log_{e} 10 = 2.3]$