JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

$250\mathrm{mL}$ of $0.5M\mathrm{NaOH}$ was added to $500\mathrm{mL}$ of $1M$ $\mathrm{HCl}$ . The number of unreacted $\mathrm{HCl}$ molecules in the solution after the complete reaction is p $\times {10}^{21}$ . Find out p

(Nearest integer) $({N}_{A}=6.022\times {10}^{23})$

Official previous-year question

Held on 20 Jul 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

We known that no. of moles $={V}_{\text{litre }}\times$ Molarity $&$ No. of millimoles $={V}_{\mathrm{ml}}\times$ Molarity so millimoles of $\mathrm{NaOH}=250\times 0.5$

$=125$

Millimoles of $\mathrm{HCl}=500\times 1=500$

Now reaction is

JEE Main 2021 Chemistry, Redox Reactions — solution figure

so millimoles of $\mathrm{HCl}$ left $=375$ Moles of $\mathrm{HCl}=375\times {10}^{-3}$

No. of $\mathrm{HCl}$ molecules $=6.022\times {10}^{23}\times 375\times {10}^{-3}$

$=225.8\times {10}^{21}$

$\simeq 226\times {10}^{21}=226$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

(Nearest integer) $({N}_{A}=6.022\times {10}^{23})$

Official previous-year question

Held on 20 Jul 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

We known that no. of moles $={V}_{\text{litre }}\times$ Molarity $&$ No. of millimoles $={V}_{\mathrm{ml}}\times$ Molarity so millimoles of $\mathrm{NaOH}=250\times 0.5$

$=125$

Millimoles of $\mathrm{HCl}=500\times 1=500$

Now reaction is

JEE Main 2021 Chemistry, Redox Reactions — solution figure

so millimoles of $\mathrm{HCl}$ left $=375$ Moles of $\mathrm{HCl}=375\times {10}^{-3}$

No. of $\mathrm{HCl}$ molecules $=6.022\times {10}^{23}\times 375\times {10}^{-3}$

$=225.8\times {10}^{21}$

$\simeq 226\times {10}^{21}=226$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2021 — Chemistry Physical Chemistry

Did you get this right?

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

Keep practicing Physical Chemistry

JEE Main 2021 — Chemistry Physical Chemistry

Did you get this right?

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

Keep practicing Physical Chemistry