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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)
Correct answer is '226'. Can you explain this answer?
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250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unr...
To determine the number of unreacted HCl molecules after the reaction, we need to calculate the number of moles of NaOH and HCl in the solution.
First, let's calculate the moles of NaOH:
Molarity (M) = moles (n) / volume (V)
0.5 M = n / 0.250 L
n = 0.5 * 0.250
n = 0.125 moles
Next, let's calculate the moles of HCl:
Molarity (M) = moles (n) / volume (V)
1 M = n / 0.500 L
n = 1 * 0.500
n = 0.500 moles
Now, we need to determine which reactant is limiting. Since the number of moles of NaOH (0.125 moles) is less than the number of moles of HCl (0.500 moles), NaOH is the limiting reactant.
The balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, all 0.125 moles of NaOH will react with 0.125 moles of HCl.
Since NaOH is the limiting reactant, the number of unreacted HCl molecules in the solution after complete reaction is 0.500 - 0.125 = 0.375 moles.
First, let's calculate the moles of NaOH:
Molarity (M) = moles (n) / volume (V)
0.5 M = n / 0.250 L
n = 0.5 * 0.250
n = 0.125 moles
Next, let's calculate the moles of HCl:
Molarity (M) = moles (n) / volume (V)
1 M = n / 0.500 L
n = 1 * 0.500
n = 0.500 moles
Now, we need to determine which reactant is limiting. Since the number of moles of NaOH (0.125 moles) is less than the number of moles of HCl (0.500 moles), NaOH is the limiting reactant.
The balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, all 0.125 moles of NaOH will react with 0.125 moles of HCl.
Since NaOH is the limiting reactant, the number of unreacted HCl molecules in the solution after complete reaction is 0.500 - 0.125 = 0.375 moles.
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Community Answer
250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unr...
We know that number of moles = Vlitre x Molarity
& No. of millimoles = Vml x Molarity
So millimoles of NaOH = 250 x 0.5
= 125
Millimoles of HCL = 500 x 1
= 500
New Reaction is
So millimoles of HCL left = 375
Moles of HCL = 375 x 10-3
No. of HCL molecules = 6.022 x 1023 x 375 x 10-3
= 225.8 x 1021
= 226 x 1021
& No. of millimoles = Vml x Molarity
So millimoles of NaOH = 250 x 0.5
= 125
Millimoles of HCL = 500 x 1
= 500
New Reaction is
So millimoles of HCL left = 375
Moles of HCL = 375 x 10-3
No. of HCL molecules = 6.022 x 1023 x 375 x 10-3
= 225.8 x 1021
= 226 x 1021
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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)Correct answer is '226'. Can you explain this answer?
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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)Correct answer is '226'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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)Correct answer is '226'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)Correct answer is '226'. Can you explain this answer?.
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)Correct answer is '226'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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)Correct answer is '226'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)Correct answer is '226'. Can you explain this answer?.
Solutions for 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)Correct answer is '226'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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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)Correct answer is '226'. Can you explain this answer?, a detailed solution for 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)Correct answer is '226'. Can you explain this answer? has been provided alongside types of 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)Correct answer is '226'. Can you explain this answer? theory, EduRev gives you an
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