A is a 4.5g of HNO3 per 1dm^3 of solution. B is a 7.0g of [XHCO3]SOH per 1dm^3 of solution. 25cm^3 of B was pipetted and titrated with A using methyl orange as an indicator. Given the titre volumes of A used to be 29.20, 29.10 and 29.30, calculate a. The average volume of A. b. The concentration of A in moldm^-3. c. The relative molecular mass of B. d. The molar mass of X
How much of this do you know how to do? Surely some of it. What do you not understand about the problem? I'll help you through it.
Can you please help me out with all the questions I have submitted 🙏
You didn't help at all. You didn't answer my questions.I could help better if I knew where you're having trouble.
a. average volumes of A, and I'm sure you already know that, is the average of the three readings.
b. (HNO3) = 4.5g/dm^3 = 4.5/63 = about 0.07 mols/dm^3 but this is a close estimate. You will need to go through ALL of these calculations since all are estimates.
c. mLA x MA = mLB x MB
Solve for M (B) = mL(A) x M(A)/mL(B) = 29.2 x 0.07/25 = about 0.08 mol/dm^3. Again, that's an estimate.
M of B = mols/dm^3 or mols = M x dm^3 = 0.08 x 1 = about 0.08. Then mols = grams/molar mass or molar mass = grams/mols = 7.0/0.08 =about 84.
d. Add all of the atomic masses of each of the elements in B, subtract from the molar mass of B and the remainder will be the molar mass of X .
Post your work if you get stuck.
NOTE: I suspect something is wrong with the numbers in your post or with the person who made up the problem for the molar mass of X is negative by my numbers and that is not possible.
After sleeping on this I had the idea to reverse the numbers; i.e., 25 cc for A and 29.10 cc for B. If I do that then M of B works out to be 0.0613 M, the molar mass of B is 114 which leaves 4 for the molar mass of X. Perhaps these numbers wee reversed by you when you posted or by the person who gave you the problem. I suspect that is what has happened.
To solve this problem, we will follow these steps:
a. Calculate the average volume of A:
1. Add the three titre volumes together: 29.20 + 29.10 + 29.30 = 87.60 cm^3
2. Divide the total by the number of measurements: 87.60 cm^3 / 3 = 29.20 cm^3
Therefore, the average volume of A is 29.20 cm^3.
b. Calculate the concentration of A in moldm^-3:
1. Calculate the amount of substance (in moles) of A used in the titration:
molar mass of HNO3 = 1*1 + 14*1 + 16*3 = 63 g/mol
mass of HNO3 used = 4.5 g
moles of HNO3 = mass / molar mass = 4.5 g / 63 g/mol = 0.0714 mol
2. Calculate the concentration of A in moldm^-3:
volume of A used = 29.20 cm^3 = 29.20/1000 dm^3 = 0.0292 dm^3
concentration of A = moles of A / volume of A = 0.0714 mol / 0.0292 dm^3
= 2.447 mol/dm^3
Therefore, the concentration of A is 2.447 moldm^-3.
c. Calculate the relative molecular mass of B:
1. The formula of B is [XHCO3]SOH, where X represents an unknown element.
Let's calculate the molar mass of [XHCO3]SOH in general:
molar mass of XHCO3 = molar mass of X + molar mass of H + molar mass of C + 3*(molar mass of O) = X + 1 + 12 + 3*16 = X + 60
molar mass of B = molar mass of XHCO3 + molar mass of SOH = (X + 60) + (32 + 16 + 1) = X + 109
2. To find the relative molecular mass of B, we need to determine the value of X.
However, we do not have enough information from the given data to calculate X. You would need additional information about B, such as the mass of B used in the reaction. Without this information, we cannot determine the relative molecular mass of B.
d. Calculate the molar mass of X:
Since we were unable to determine the relative molecular mass of B in the previous step, we cannot calculate the molar mass of X.
Additional information is required to find the molar mass of X.