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A solution of HNO3 is standardized by reaction with pure sodium carbonate. 2HNO3 + Na2CO3 ⟶ 2NaNO3 + H2O + CO2. A volume of 25.36 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.8311 ± 0.0007g of Na2CO3, (FM 105.988 ± 0.001 g/mol). Find the molarity of the HNO3 solution and its absolute uncertainty
THE ANSWERS ARE:
MOLARITY= 0.3092mol/l
ABSOLUTE UNCERTAINTY= 0.000873
Explanation:
CALCULATION FOR MOLARITY:
Molarity= gram mole of solute / liters of solution
Where;
Mole= mass/molar mass
mole= 0.8311g/105.988g/mol= 0.0078515mol
MOLARITY= 0.0078415mol/25.36ml = 0.0003092mol/ml = 0.3092mol/l
CALCULATION FOR ABSOLUTE UNCERTAINTY:
Uncertainty (u) =√({0.05/25.36}^2 + {0.001/105.988}^2 + {0.0007/0.8311}^2) × Molarity
Solving brackets gives
(0.00197161+0.00000943503+0.00084226) ×Molarity
Adding up gives
0.002823×Molarity
Therefore;
ABSOLUTE UNCERTAINTY= 0.002823×0.3092= 0.000873