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 and its absolute uncertainty, we can use the formula:
Molarity (M) = (moles of solute)/(volume of solution in liters)
First, let's calculate the number of moles of Na2CO3:
Moles of Na2CO3 = Mass of Na2CO3 / Formula Mass of Na2CO3
Given the mass of Na2CO3 is 0.8896 ± 0.0008 g, and the formula mass of Na2CO3 is 105.988 ± 0.001 g/mol, we can substitute these values into the formula:
Moles of Na2CO3 = 0.8896 g / 105.988 g/mol
Moles of Na2CO3 = 0.0084 mol (rounded to four significant figures since 0.8896 has four significant figures)
Next, let's calculate the molarity of HNO3:
Molarity of HNO3 = Moles of HNO3 / Volume of HNO3 in liters
Since the volume of HNO3 used was 29.62 ± 0.05 mL, let's convert it to liters:
Volume of HNO3 = 29.62 mL * (1 L / 1000 mL) = 0.02962 L (rounded to five significant figures)
Now, we substitute the moles of Na2CO3 (which is the same as moles of HNO3 since the reaction is in a 1:1 ratio) and the volume of HNO3 into the formula:
Molarity of HNO3 = 0.0084 mol / 0.02962 L
Molarity of HNO3 = 0.2838 M (rounded to four significant figures since 0.0084 has four significant figures)
Therefore, the molarity of HNO3 is 0.2838 M.
To find the absolute uncertainty in the molarity, we need to consider the uncertainties in both the mass of Na2CO3 and the volume of HNO3. The absolute uncertainty is determined by taking the maximum deviation from the calculated value.
For the mass of Na2CO3:
Absolute uncertainty = 0.0008 g
For the volume of HNO3:
Absolute uncertainty = 0.05 mL * (1 L / 1000 mL) = 0.00005 L
Now, we can calculate the absolute uncertainty in the molarity of HNO3 using the formula:
Absolute uncertainty in molarity = Molarity * (Absolute uncertainty in mass / Mass of Na2CO3 + Absolute uncertainty in volume / Volume of HNO3)
Substituting the values:
Absolute uncertainty in molarity = 0.2838 M * (0.0008 g / 0.8896 g + 0.00005 L / 0.02962 L)
Absolute uncertainty in molarity = 0.00080 M (rounded to five significant figures)
So, the molarity of HNO3 is approximately 0.2838 M with an absolute uncertainty of 0.00080 M.
To find the molarity of HNO3 and its absolute uncertainty, we can use the formula:
Molarity (M) = (amount of solute in moles) / (volume of solution in liters)
Step 1: Calculate the number of moles of Na2CO3 using the given mass and formula mass.
Number of moles of Na2CO3 = (mass of Na2CO3) / (formula mass of Na2CO3)
= (0.8896 ± 0.0008 g) / (105.988 ± 0.001 g/mol)
Let's first calculate the value of the number of moles of Na2CO3:
Number of moles of Na2CO3 = 0.8896 g / 105.988 g/mol
≈ 0.00838 mol
Step 2: Determine the absolute uncertainty in the number of moles of Na2CO3.
Absolute uncertainty in moles of Na2CO3 = (Absolute uncertainty in mass of Na2CO3) / (formula mass of Na2CO3)
[Note: We are given the absolute uncertainty in mass, but the formula mass has a negligible uncertainty.]
Absolute uncertainty in moles of Na2CO3 = 0.0008 g / 105.988 g/mol
≈ 7.56 x 10^-6 mol
Step 3: Calculate the volume of HNO3 solution in liters.
Volume of HNO3 solution = 29.62 ± 0.05 mL
= (29.62 ± 0.05) / 1000 L
Let's first calculate the value of the volume of HNO3 solution:
Volume of HNO3 solution = 29.62 mL / 1000 L
≈ 0.02962 L
Step 4: Calculate the molarity of HNO3.
Molarity of HNO3 = (number of moles of Na2CO3) / (volume of HNO3 solution)
= (0.00838 mol) / (0.02962 L)
Let's first calculate the value of the molarity of HNO3:
Molarity of HNO3 = 0.2827 M
Step 5: Determine the absolute uncertainty in the molarity of HNO3.
Absolute uncertainty in molarity of HNO3 = | (Absolute uncertainty in number of moles of Na2CO3) / (volume of HNO3 solution) |
= | (7.56 x 10^-6 mol) / (0.02962 L) |
Let's first calculate the value of the absolute uncertainty in the molarity of HNO3:
Absolute uncertainty in molarity of HNO3 = | 2.5567 x 10^-4 M |