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Question: Answered & Verified by Expert
250 mL of 0.5M NaOH was added to 500 mL of 1M HCl. The number of unreacted HCl molecules in the solution after the complete reaction is p×1021. Find out p
(Nearest integer) NA=6.022×1023
ChemistryRedox ReactionsJEE MainJEE Main 2021 (20 Jul Shift 1)
Solution:
2665 Upvotes Verified Answer
The correct answer is: 226

We known that no. of moles =Vlitre × Molarity & No. of millimoles =Vml× Molarity so millimoles of NaOH=250×0.5
=125
Millimoles of HCl=500×1=500
Now reaction is

so millimoles of HCl left =375 Moles of HCl=375×10-3
No. of HCl molecules =6.022×1023×375×10-3
=225.8×1021
226×1021=226

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