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1.0 g of pure calcium carbonate was found to require 50 ml of dilute HCl for complete reaction. The strength of HCl solution is given by
- a)4 N
- b)2 N
- c)0.4 N
- d)0.2 N
Correct answer is option 'C'. Can you explain this answer?
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1.0 g of pure calcium carbonate was found to require 50 ml of dilute H...
Calculation of the strength of HCl solution
Given data:
Mass of calcium carbonate = 1.0 g
Volume of HCl used = 50 ml
Step 1: Calculation of the moles of calcium carbonate
Moles = Mass/Molar mass
Molar mass of calcium carbonate = 100.08 g/mol
Moles of calcium carbonate = 1.0/100.08 = 0.00999 mol
Step 2: Calculation of the moles of HCl used
Moles of HCl = Molarity × Volume (in liters)
Volume of HCl used = 50 ml = 0.05 L
Let the molarity of HCl be 'x'
Moles of HCl = x × 0.05
Step 3: Calculation of the molarity of HCl solution
The balanced chemical equation for the reaction between calcium carbonate and HCl is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, it is clear that 1 mole of calcium carbonate reacts with 2 moles of HCl. Therefore,
Moles of HCl used = 2 × Moles of calcium carbonate
x × 0.05 = 2 × 0.00999
x = (2 × 0.00999)/0.05
x = 0.3996 M
Step 4: Conversion of molarity to normality
Normality = Molarity × Basicity
For HCl, Basicity = 1 (as it donates 1 H+ ion)
Therefore, Normality = Molarity × 1
Normality = 0.3996 N
Hence, the strength of HCl solution is 0.4 N.
Given data:
Mass of calcium carbonate = 1.0 g
Volume of HCl used = 50 ml
Step 1: Calculation of the moles of calcium carbonate
Moles = Mass/Molar mass
Molar mass of calcium carbonate = 100.08 g/mol
Moles of calcium carbonate = 1.0/100.08 = 0.00999 mol
Step 2: Calculation of the moles of HCl used
Moles of HCl = Molarity × Volume (in liters)
Volume of HCl used = 50 ml = 0.05 L
Let the molarity of HCl be 'x'
Moles of HCl = x × 0.05
Step 3: Calculation of the molarity of HCl solution
The balanced chemical equation for the reaction between calcium carbonate and HCl is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, it is clear that 1 mole of calcium carbonate reacts with 2 moles of HCl. Therefore,
Moles of HCl used = 2 × Moles of calcium carbonate
x × 0.05 = 2 × 0.00999
x = (2 × 0.00999)/0.05
x = 0.3996 M
Step 4: Conversion of molarity to normality
Normality = Molarity × Basicity
For HCl, Basicity = 1 (as it donates 1 H+ ion)
Therefore, Normality = Molarity × 1
Normality = 0.3996 N
Hence, the strength of HCl solution is 0.4 N.
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1.0 g of pure calcium carbonate was found to require 50 ml of dilute HCl for complete reaction. The strength of HCl solution is given bya)4 Nb)2 Nc)0.4 Nd)0.2 NCorrect answer is option 'C'. Can you explain this answer?
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