A volume of 25.36 ± 0.05 mL 25.36±0.05 mL of HNO 3 HNO3 solution was required for complete reaction with 0.8311 ± 0.0007 g 0.8311±0.0007 g of Na 2 CO 3 Na2CO3, (FM 105.988 ± 0.001 g/mol 105.988±0.001 g/mol). Find the molarity of the HNO 3 HNO3 solution and its absolute uncertainty.

MOLARITY= 0.3092mol/l

ABSOLUTE UNCERTAINTY= 0.000873

Explanation:

The equation of reaction is

2HNO3 + Na2CO3 ⟶ 2NaNO3 + H2O + CO2.

QUESTION1: CALCULATION FOR MOLARITY;

Molarity= gram mole of solute ÷ liters of solution

Where;

Mole of solute= mass ÷ molar mass

Therefore;

Mole of solute= 0.8311g ÷ 105.988g/mol= 0.0078515mol

MOLARITY= 0.0078415mol ÷ 25.36ml = 0.0003092mol/ml = 0.3092mol/l

This is the Molarity of the solution

QUESTION2: 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

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol