JEE Main 2021 — Some 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)
(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

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.
So, millimoles of NaOH = 250 0.5 = 125
Millimoles of HCl = 500 1 = 500
Now, reaction is

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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