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0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutralization. Molecular wt. of acid is:
- a)32
- b)64
- c)128
- d)256
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutrali...
At neutralisation,
25 ml N/10 NaOH ≡≡25 ml N/10 acid
Now,
25 ml N/10 acid = 0.16 gm
So,
1000 ml N/10 acid = (0.16/25) x 1000
= 6.4 g
Therefore,
1000 ml N acid = 6.4 x 10 = 64 g
Thus, the equivalent weight of acid = 64 g
Since the acid is dibasic,
the molecular weight of acid = 64 x 2 = 128 g
Most Upvoted Answer
0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutrali...
To find the molecular weight of the dibasic acid, we need to use the concept of equivalent weight and the given information about the volume and concentration of NaOH used for neutralization.
Here's how we can solve it step by step:
1. Write the balanced chemical equation for the reaction between the dibasic acid and NaOH. Let's assume the acid is H2A:
H2A + 2NaOH → Na2A + 2H2O
2. From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of dibasic acid (H2A). Therefore, the equivalent weight of the acid is half of the molecular weight.
3. Convert the given volume of NaOH (25 mL) to moles using the molarity (N/10) of NaOH. The molarity is defined as the number of moles of solute per liter of solution, so we need to convert the volume to liters:
25 mL = 25/1000 = 0.025 L
Moles of NaOH = Molarity × Volume = (N/10) × 0.025 L
4. Since 2 moles of NaOH react with 1 mole of dibasic acid, the moles of dibasic acid can be calculated using the stoichiometry of the balanced equation:
Moles of dibasic acid (H2A) = 0.5 × Moles of NaOH
5. Now, we can calculate the molecular weight of the dibasic acid using the given mass (0.16 g) and the moles of dibasic acid:
Molecular weight = Mass/Moles
Substituting the values:
Molecular weight = 0.16 g / (0.5 × Moles of NaOH)
6. To find the molecular weight of the acid, we need to calculate the moles of NaOH first:
Moles of NaOH = (N/10) × 0.025 L
Substituting the value of Moles of NaOH in the equation for Molecular weight:
Molecular weight = 0.16 g / (0.5 × (N/10) × 0.025 L)
7. Simplifying the equation:
Molecular weight = (0.16 g × 10 × 0.025 L) / (0.5 × N)
Molecular weight = 0.16 g × 5 × 0.025 L / N
Molecular weight = 0.02 g × L / N
8. Comparing the equation with the standard formula for equivalent weight:
Molecular weight = Equivalent weight × Valency
From the equation, we can see that the equivalent weight is 0.02 g/L and the valency is 2 because it is a dibasic acid.
9. Finally, we can calculate the molecular weight:
Molecular weight = Equivalent weight × Valency = 0.02 g/L × 2 = 0.04 g/L
To convert grams per liter to grams per mole, we need to multiply by the molar volume (22.4 L/mol) at STP (standard temperature and pressure):
Molecular weight = 0.04 g/L × 22.4 L/mol = 0.896 g/mol
Therefore, the molecular weight of the dibasic acid is approximately 0.896 g/mol,
Here's how we can solve it step by step:
1. Write the balanced chemical equation for the reaction between the dibasic acid and NaOH. Let's assume the acid is H2A:
H2A + 2NaOH → Na2A + 2H2O
2. From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of dibasic acid (H2A). Therefore, the equivalent weight of the acid is half of the molecular weight.
3. Convert the given volume of NaOH (25 mL) to moles using the molarity (N/10) of NaOH. The molarity is defined as the number of moles of solute per liter of solution, so we need to convert the volume to liters:
25 mL = 25/1000 = 0.025 L
Moles of NaOH = Molarity × Volume = (N/10) × 0.025 L
4. Since 2 moles of NaOH react with 1 mole of dibasic acid, the moles of dibasic acid can be calculated using the stoichiometry of the balanced equation:
Moles of dibasic acid (H2A) = 0.5 × Moles of NaOH
5. Now, we can calculate the molecular weight of the dibasic acid using the given mass (0.16 g) and the moles of dibasic acid:
Molecular weight = Mass/Moles
Substituting the values:
Molecular weight = 0.16 g / (0.5 × Moles of NaOH)
6. To find the molecular weight of the acid, we need to calculate the moles of NaOH first:
Moles of NaOH = (N/10) × 0.025 L
Substituting the value of Moles of NaOH in the equation for Molecular weight:
Molecular weight = 0.16 g / (0.5 × (N/10) × 0.025 L)
7. Simplifying the equation:
Molecular weight = (0.16 g × 10 × 0.025 L) / (0.5 × N)
Molecular weight = 0.16 g × 5 × 0.025 L / N
Molecular weight = 0.02 g × L / N
8. Comparing the equation with the standard formula for equivalent weight:
Molecular weight = Equivalent weight × Valency
From the equation, we can see that the equivalent weight is 0.02 g/L and the valency is 2 because it is a dibasic acid.
9. Finally, we can calculate the molecular weight:
Molecular weight = Equivalent weight × Valency = 0.02 g/L × 2 = 0.04 g/L
To convert grams per liter to grams per mole, we need to multiply by the molar volume (22.4 L/mol) at STP (standard temperature and pressure):
Molecular weight = 0.04 g/L × 22.4 L/mol = 0.896 g/mol
Therefore, the molecular weight of the dibasic acid is approximately 0.896 g/mol,
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Community Answer
0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutrali...
Look here the acid is dibasic so it's n factor is 2 now the molarity of acid is defined as no of moles per litrr so molarity of acid is 0.16/M *1L (M is the molecular wt of acid ) so it's normality is molarity *nfactor i.e .16/M*1L *2. applying. N1V1=N2V2 1/10 * 25/1000 L = .16/M*1L *2 * 1L on solving we get M= 128g
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0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutralization. Molecular wt. of acid is:a)32b)64c)128d)256Correct answer is option 'C'. Can you explain this answer?
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