Test Index

JEE Advanced 2021 Paper 1

Show Para Hide Para

Paragraph for next 2 questions

For the following reaction scheme, percentage yields are given along the arrow:

x g

and

y g

are mass of

R

and

U

, respectively.

(Use : Molar mass (in

{g} {mol}^{-1}

) of

{H}, {C}

and

{O}

as 1,12 and 16 , respectively)

Section: Chemistry

© examsnet.com

Question : 25 of 57

Marks: +1, -0

The value of

y

[JEE Adv 2021 P1]

Your Answer:

Solution:

\mathrm{CH}_3 \underset{(0.1)\text{mole}}{\underset{(P)}{\mathrm{C}-\mathrm{C}}} \equiv \mathrm{CH} \xrightarrow[\text{Kucherov reaction}]{\frac{\mathrm{Hg}^{2+} / \mathrm{H}^{+}}{333\,\text{K}, 100\%}}

From

0.1\,\text{mol}

of

\mathrm{P}, 0.032\,\text{mol}

of

U

is produced.

So, the value of

y = 0.032 \times 100 = 3.2\,\text{g}

© examsnet.com

Go to Question:

Prev Question

Next Question