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JEE Main 2021Some Basic Concepts Of Chemistry Question with Solution

From: JEE Main 2021 (Online) 20th July Morning Shift

Question

250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ___________ 1021. (Nearest integer)

(NA = 6.022 1023)

Enter your answer

Show full solutionCorrect answer: 226
Correct answer
226

Step-by-step explanation

We know that, number of moles = VL molarity and number of millimoles = VmL molarity

So, millimoles of NaOH = 250 0.5 = 125

Millimoles of HCl = 500 1 = 500

Now, reaction is

JEE Main 2021 (Online) 20th July Morning Shift Chemistry - Some Basic Concepts of Chemistry Question 142 English Explanation
125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375

Moles of HCl = 375 103

Number of HCl molecules

= Avogadro's constant (NA) moles of HCl

= 6.022 1023 375 103

= 225.8 1021 = 226 1021

Therefore, answer is 226.

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About this question

This is a previous-year question from JEE Main 2021, covering the Some Basic Concepts Of Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.