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50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
- a)10 g
- b)4 g
- c)20 g
- d)80 g
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hyd...
2 × 50 × 0.5 = 25 × M
⇒ M = 2
∴ Moles of NaOH in 50 mL 
∴ Weight = 4 grams
Most Upvoted Answer
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hyd...
To find the amount of NaOH in 50 mL of the given sodium hydroxide solution, we can use the concept of stoichiometry.
Stoichiometry is a mathematical relationship between the amounts of reactants and products in a chemical reaction. In this case, the reaction is between oxalic acid (H2C2O4) and sodium hydroxide (NaOH), which can be represented by the balanced chemical equation:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we can see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.
First, let's calculate the number of moles of oxalic acid in 50 mL of 0.5 M solution.
Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, the number of moles of oxalic acid (H2C2O4) in 50 mL of 0.5 M solution can be calculated as follows:
Number of moles = Molarity × Volume (in liters)
Volume of solution = 50 mL = 50/1000 L = 0.05 L
Number of moles of H2C2O4 = 0.5 M × 0.05 L = 0.025 moles
According to the balanced equation, 1 mole of H2C2O4 reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH required to neutralize the given amount of H2C2O4 is twice the number of moles of H2C2O4:
Number of moles of NaOH = 2 × 0.025 moles = 0.05 moles
Finally, let's calculate the amount of NaOH in 50 mL of the given sodium hydroxide solution.
The molar mass of NaOH is 23 + 16 + 1 = 40 g/mol.
Amount of NaOH = Number of moles × Molar mass
Amount of NaOH = 0.05 moles × 40 g/mol = 2 g
Therefore, the amount of NaOH in 50 mL of the given sodium hydroxide solution is 2 g, which corresponds to option B.
Stoichiometry is a mathematical relationship between the amounts of reactants and products in a chemical reaction. In this case, the reaction is between oxalic acid (H2C2O4) and sodium hydroxide (NaOH), which can be represented by the balanced chemical equation:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we can see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.
First, let's calculate the number of moles of oxalic acid in 50 mL of 0.5 M solution.
Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, the number of moles of oxalic acid (H2C2O4) in 50 mL of 0.5 M solution can be calculated as follows:
Number of moles = Molarity × Volume (in liters)
Volume of solution = 50 mL = 50/1000 L = 0.05 L
Number of moles of H2C2O4 = 0.5 M × 0.05 L = 0.025 moles
According to the balanced equation, 1 mole of H2C2O4 reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH required to neutralize the given amount of H2C2O4 is twice the number of moles of H2C2O4:
Number of moles of NaOH = 2 × 0.025 moles = 0.05 moles
Finally, let's calculate the amount of NaOH in 50 mL of the given sodium hydroxide solution.
The molar mass of NaOH is 23 + 16 + 1 = 40 g/mol.
Amount of NaOH = Number of moles × Molar mass
Amount of NaOH = 0.05 moles × 40 g/mol = 2 g
Therefore, the amount of NaOH in 50 mL of the given sodium hydroxide solution is 2 g, which corresponds to option B.
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50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution isa)10 gb)4gc)20 gd)80 gCorrect answer is option 'B'. Can you explain this answer?
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