![JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Some_Basic_Concepts_of_Chemistry/648b5a69417cc3fb48d66fb2/question_1__q_648b5a69417cc3fb48d66fb2__cdn-question-pool.getmarks.app__c3306daf-dba5-4849-b7de-c89f4531253c-image__3f1ede3b00_final_ppt_sync.png) Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass : $C:12.0u,H:1.0u$ $N:14.0u,O:16.0u,\mathrm{Cl}:35.5u)$

![JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 1](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Some_Basic_Concepts_of_Chemistry/648b5a69417cc3fb48d66fb2/solution_1__q_648b5a69417cc3fb48d66fb2__cdn-question-pool.getmarks.app__fb734d1d-1107-4779-92cb-927c7a2c6086-image__c15eabaf7f_final_ppt_sync.png) $1\mathrm{mole}1\mathrm{mole}1\mathrm{mole}$ $=140.5\mathrm{gm}=169\mathrm{gm}=273\mathrm{gm}$ $\therefore 0.140\mathrm{gm}\frac{169}{140.5}\times 0.140$ $L.R.=0.168\mathrm{gm}<0.388\mathrm{gm}$ excess $\therefore$ Theoretical amount of given product formed $=\frac{273}{140.5}\times 0.140=0.272\mathrm{gm}$ But its actual amount formed is $0.210\mathrm{gm}$. Hence, the percentage yield of product. $=\frac{0.210}{0.272}\times 100=77.20\approx 77$ OR ![JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 2](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Some_Basic_Concepts_of_Chemistry/648b5a69417cc3fb48d66fb2/solution_2__q_648b5a69417cc3fb48d66fb2__cdn-question-pool.getmarks.app__e74e35f6-bd7c-4eb2-9563-72448541458a-image__fc4ceb45e2_final_ppt_sync.png) Mole of $\mathrm{Ph}-\mathrm{CoCl}=\frac{0.140}{140}={10}^{-3}\mathrm{mol}$ ![JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 3](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Some_Basic_Concepts_of_Chemistry/648b5a69417cc3fb48d66fb2/solution_3__q_648b5a69417cc3fb48d66fb2__cdn-question-pool.getmarks.app__c54d2ae9-6bc1-4e88-aee0-3a36ff488c05-image__9999b130ef_final_ppt_sync.png) by mol-mol analysis $={10}^{-3}$ mol. Theoritical mass of product $={10}^{-3}\times 273=$ $273\times {10}^{-3}g$ Observed mass of product $=210\times {10}^{-3}g$ $%$ yield of product $=\frac{210\times {10}^{-3}}{273\times {10}^{-3}}\times 100=76.9%=77$

JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — question figure

Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass : $C:12.0u,H:1.0u$ $N:14.0u,O:16.0u,\mathrm{Cl}:35.5u)$

Official previous-year question

Held on 17 Mar 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 1

$1\mathrm{mole}1\mathrm{mole}1\mathrm{mole}$

$=140.5\mathrm{gm}=169\mathrm{gm}=273\mathrm{gm}$

$\therefore 0.140\mathrm{gm}\frac{169}{140.5}\times 0.140$

$L.R.=0.168\mathrm{gm}<0.388\mathrm{gm}$

excess

$\therefore$ Theoretical amount of given product formed

$=\frac{273}{140.5}\times 0.140=0.272\mathrm{gm}$

But its actual amount formed is $0.210\mathrm{gm}$ . Hence, the percentage yield of product.

$=\frac{0.210}{0.272}\times 100=77.20\approx 77$

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 2

Mole of $\mathrm{Ph}-\mathrm{CoCl}=\frac{0.140}{140}={10}^{-3}\mathrm{mol}$

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 3

by mol-mol analysis $={10}^{-3}$ mol.

Theoritical mass of product $={10}^{-3}\times 273=$ $273\times {10}^{-3}g$

Observed mass of product $=210\times {10}^{-3}g$

$%$ yield of product $=\frac{210\times {10}^{-3}}{273\times {10}^{-3}}\times 100=76.9%=77$

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JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — question figure

Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass : $C:12.0u,H:1.0u$ $N:14.0u,O:16.0u,\mathrm{Cl}:35.5u)$

Official previous-year question

Held on 17 Mar 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 1

$1\mathrm{mole}1\mathrm{mole}1\mathrm{mole}$

$=140.5\mathrm{gm}=169\mathrm{gm}=273\mathrm{gm}$

$\therefore 0.140\mathrm{gm}\frac{169}{140.5}\times 0.140$

$L.R.=0.168\mathrm{gm}<0.388\mathrm{gm}$

excess

$\therefore$ Theoretical amount of given product formed

$=\frac{273}{140.5}\times 0.140=0.272\mathrm{gm}$

But its actual amount formed is $0.210\mathrm{gm}$ . Hence, the percentage yield of product.

$=\frac{0.210}{0.272}\times 100=77.20\approx 77$

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 2

Mole of $\mathrm{Ph}-\mathrm{CoCl}=\frac{0.140}{140}={10}^{-3}\mathrm{mol}$

JEE Main 2021 Chemistry, Some Basic Concepts of Chemistry — solution figure 3

by mol-mol analysis $={10}^{-3}$ mol.

Theoritical mass of product $={10}^{-3}\times 273=$ $273\times {10}^{-3}g$

Observed mass of product $=210\times {10}^{-3}g$

$%$ yield of product $=\frac{210\times {10}^{-3}}{273\times {10}^{-3}}\times 100=76.9%=77$

JEE Main 2021 — Chemistry Physical Chemistry

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Solution

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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