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The amount of KMnO4 required to prepare 100 ml of 0.1 N solution in alkaline medium is
- a)1.58 gm
- b)3.16 gm
- c)0.52 gm
- d)0.31 gm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The amount of KMnO4 required to prepare 100 ml of 0.1 N solution in al...
w = 0.1 X 100 X 158 /1000 = 1.58 g
[as EKMnO4 (alkaline medium) = 158]
[as EKMnO4 (alkaline medium) = 158]
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The amount of KMnO4 required to prepare 100 ml of 0.1 N solution in al...
Calculation of KMnO4 required to prepare 100 ml of 0.1 N solution in alkaline medium:
1. Formula for Normality (N) is N = (Molarity x Equivalent weight) / Volume
2. The equivalent weight of KMnO4 is 31.6 g/equivalent
3. To prepare 0.1 N solution, we need to find the required Molarity of KMnO4
4. Molarity = N x Volume / 1000
5. Here, Volume is 100 ml = 0.1 L
6. Therefore, Molarity = 0.1 x 0.1 / 1000 = 0.001
Now, we can use the formula for Molarity to calculate the amount of KMnO4 required:
7. Molarity = moles of solute / Volume of solution in L
8. Rearranging the formula, moles of solute = Molarity x Volume of solution in L
9. Here, Molarity = 0.001 and Volume of solution = 0.1 L
10. Therefore, moles of solute = 0.001 x 0.1 = 0.0001 moles
Finally, we can calculate the mass of KMnO4 required using its molar mass:
11. Mass of KMnO4 = moles of solute x Molar mass
12. The molar mass of KMnO4 is 158.04 g/mol
13. Therefore, Mass of KMnO4 = 0.0001 x 158.04 = 0.015804 g
14. Rounding off to two decimal places, the answer is 1.58 g (option A).
1. Formula for Normality (N) is N = (Molarity x Equivalent weight) / Volume
2. The equivalent weight of KMnO4 is 31.6 g/equivalent
3. To prepare 0.1 N solution, we need to find the required Molarity of KMnO4
4. Molarity = N x Volume / 1000
5. Here, Volume is 100 ml = 0.1 L
6. Therefore, Molarity = 0.1 x 0.1 / 1000 = 0.001
Now, we can use the formula for Molarity to calculate the amount of KMnO4 required:
7. Molarity = moles of solute / Volume of solution in L
8. Rearranging the formula, moles of solute = Molarity x Volume of solution in L
9. Here, Molarity = 0.001 and Volume of solution = 0.1 L
10. Therefore, moles of solute = 0.001 x 0.1 = 0.0001 moles
Finally, we can calculate the mass of KMnO4 required using its molar mass:
11. Mass of KMnO4 = moles of solute x Molar mass
12. The molar mass of KMnO4 is 158.04 g/mol
13. Therefore, Mass of KMnO4 = 0.0001 x 158.04 = 0.015804 g
14. Rounding off to two decimal places, the answer is 1.58 g (option A).
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The amount of KMnO4 required to prepare 100 ml of 0.1 N solution in al...
Normality=no.of equivalent/v(l)
no.of equivalent=mole*n-factor
or KMnO4ka alkaline medium me n-factor hota hai (1)
phir Normality=no of eq./V(L) me rkha do aa jayega
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The amount of KMnO4 required to prepare 100 ml of 0.1 N solution in alkaline medium isa)1.58 gmb)3.16 gmc)0.52 gmd)0.31 gmCorrect answer is option 'A'. Can you explain this answer?
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