Answer:
Molarity of HNO₃ = 0.6379 ± 0.0020 M
Step-by-step explanation:
Hi!
The equation of the reaction is as follows:
Na₂CO₃ + 2HNO₃ → H₂CO₃ + 2NaNO₃
The number of moles of Na₂CO₃ present in the reaction is:
moles Na₂CO₃ = 0.8359 ± 0.0007 g / 105.988 ± 0.001 g/mol
First, let´s convert the absolute uncertainties in relative ones.
0.8359 ± 0.0007 g = 0.8359 ± ( (0.0007 g/ 0.8359 g) * 100%)
=0.8359g ± 0.08%
105.988 ± 0.001 g = 105.988 ± ((0.001 g/mol / 105.988 g/mol) * 100%)
= 105.988 g/mol ± 9×10⁻⁴%
Then:
moles Na₂CO₃ = (0.8359g ± 0.08%) / (105.988 g/mol ± 9×10⁻⁴%)
moles Na₂CO₃ = (0.8359 g / 105.988 g/mol) ± (0.08% + 9×10⁻⁴%)
moles Na₂CO₃ = 7.887 × 10⁻³ mol ± 0.08%
From the equation, we know that 2 moles of HNO₃ react with 1 mol of Na₂CO₃. Then, the number of moles of HNO₃ present in the reaction is:
moles of HNO₃ = 2 * moles of Na₂CO₃ = 2 * 7.887 × 10⁻³ mol ± 0.08%
moles of HNO₃ = 0.01577 mol ± 0.08%
This number of moles was present in 24.72 ml ± 0.2%. Then, in 1 l there will be:
moles of HNO₃ in 1l = 0.01577 mol ± 0.08% * 1000 ml / 24.72 ml ± 0.2%
moles of HNO₃ in 1l = 0.6379 mol ± (0.08% + 0.2%)
moles of HNO₃ in 1l = 0.6379 mol ± 0.3%
Converting the relative uncertainty to absolute one:
0.3% * 0.6379 mol / 100% = 2 × 10⁻³ mol
Then, the molarity of HNO₃ will be:
Molarity of HNO₃ = 0.6379 ± 0.0020 M