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Question

For the inversion of cane sugar, (C12H22O11) obeying I order following data were obtained.

Time (min.)010
Angle of rotation of solution(degree)+202.510

What will be rate constant in min1?
(ln2=0.7)

A
0.7
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B
0.14
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C
0.21
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D
0.07
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Solution

The correct option is A 0.14

Since inversion of sugar is first order reaction, value of k is given by:
k=1tlnrorrtr
k=110ln20(10)(2.5)(10)
k=ln4=0.14


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