JEE MAIN - Chemistry (2021 - 20th July Morning Shift - No. 13)
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 ___________ $$\times$$ 1021. (Nearest integer)
(NA = 6.022 $$\times$$ 1023)
(NA = 6.022 $$\times$$ 1023)
Answer
226
Explanation
We know that, number of moles = VL $$\times$$ molarity and number of millimoles = VmL $$\times$$ molarity
So, millimoles of NaOH = 250 $$\times$$ 0.5 = 125
Millimoles of HCl = 500 $$\times$$ 1 = 500
Now, reaction is
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125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 $$\times$$ 10$$-$$3
Number of HCl molecules
= Avogadro's constant (NA) $$\times$$ moles of HCl
= 6.022 $$\times$$ 1023 $$\times$$ 375 $$\times$$ 10$$-$$3
= 225.8 $$\times$$ 1021 = 226 $$\times$$ 1021
Therefore, answer is 226.
So, millimoles of NaOH = 250 $$\times$$ 0.5 = 125
Millimoles of HCl = 500 $$\times$$ 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 $$\times$$ 10$$-$$3
Number of HCl molecules
= Avogadro's constant (NA) $$\times$$ moles of HCl
= 6.022 $$\times$$ 1023 $$\times$$ 375 $$\times$$ 10$$-$$3
= 225.8 $$\times$$ 1021 = 226 $$\times$$ 1021
Therefore, answer is 226.
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