A volume of 29.62 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.8896 ± 0.0008 g of Na2CO3, (FM 105.988 ± 0.001).
Find the molarity of the HNO3 and its absolute uncertainty.
I have the molarity of HNO3 = 0.5667
To find the molarity of HNO3, you can use the following formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, let's calculate the moles of Na2CO3 using its mass and molar mass. The molar mass of Na2CO3 is 105.988 g/mol.
Mass of Na2CO3 (g) = 0.8896 g
Molar mass of Na2CO3 (g/mol) = 105.988 g/mol
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
Moles of Na2CO3 = 0.8896 g / 105.988 g/mol
Next, we need to convert the volume of the HNO3 solution from mL to liters.
Volume of HNO3 solution = 29.62 ± 0.05 mL
Volume of HNO3 solution (L) = 29.62 mL / 1000 mL/L
Volume of HNO3 solution (L) = 0.02962 L
Now, we can use the obtained values to calculate the molarity of HNO3.
Molarity (M) = (moles of Na2CO3) / (volume of HNO3 solution in liters)
Molarity (M) = (0.8896 g / 105.988 g/mol) / 0.02962 L
Calculating this expression results in a molarity of approximately 0.5667 M, as you mentioned.
To calculate the absolute uncertainty in the molarity, we need to consider the uncertainties in both the mass of Na2CO3 and the volume of the HNO3 solution.
Absolute uncertainty in the molarity can be calculated using the following formula:
Absolute uncertainty in Molarity = Molarity × (Relative uncertainty in mass + Relative uncertainty in volume)
Let's calculate the relative uncertainties first:
Relative uncertainty in mass = Uncertainty in mass / Mass of Na2CO3
Relative uncertainty in mass = 0.0008 g / 0.8896 g
Relative uncertainty in volume = Uncertainty in volume / Volume of HNO3 solution
Relative uncertainty in volume = 0.05 mL / 29.62 mL
Now we can calculate the absolute uncertainty in the molarity:
Absolute uncertainty in Molarity = 0.5667 M × ( (0.0008 g / 0.8896 g) + (0.05 mL / 29.62 mL) )
Evaluating this expression gives you the absolute uncertainty in the molarity.