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The weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution is
- a)4 g
- b)1 g
- c)2 g
- d)10 g
Correct answer is option 'B'. Can you explain this answer?
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The weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution ...
Calculation of Molarity and Molecular Weight:
To find the weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution, we need to calculate the molarity of NaOH and its molecular weight.
Molarity (M) = Number of moles of solute / Volume of solution in liters
Here, the volume of solution is given in cm3, so we need to convert it into liters.
250 cm3 = 250/1000 L = 0.25 L
Now, we can find the number of moles of NaOH:
0.1 N solution means 0.1 moles of NaOH per liter of solution.
So, for 0.25 L of solution, the number of moles of NaOH will be:
0.1 x 0.25 = 0.025 moles
Molecular weight (MW) = Mass of solute / Number of moles of solute
We can rearrange this formula to find the mass of solute:
Mass of solute = Molecular weight x Number of moles of solute
The molecular weight of NaOH is:
Na (23) + O (16) + H (1) = 40 g/mol
So, the mass of NaOH required will be:
0.025 x 40 = 1 g
Therefore, the correct answer is option (B) 1 g.
To find the weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution, we need to calculate the molarity of NaOH and its molecular weight.
Molarity (M) = Number of moles of solute / Volume of solution in liters
Here, the volume of solution is given in cm3, so we need to convert it into liters.
250 cm3 = 250/1000 L = 0.25 L
Now, we can find the number of moles of NaOH:
0.1 N solution means 0.1 moles of NaOH per liter of solution.
So, for 0.25 L of solution, the number of moles of NaOH will be:
0.1 x 0.25 = 0.025 moles
Molecular weight (MW) = Mass of solute / Number of moles of solute
We can rearrange this formula to find the mass of solute:
Mass of solute = Molecular weight x Number of moles of solute
The molecular weight of NaOH is:
Na (23) + O (16) + H (1) = 40 g/mol
So, the mass of NaOH required will be:
0.025 x 40 = 1 g
Therefore, the correct answer is option (B) 1 g.
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The weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution ...
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The weight of pure NaOH required to prepare 250 cm3 of 0.1 N solution isa)4 gb)1 gc)2 gd)10 gCorrect answer is option 'B'. Can you explain this answer?
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