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
4 answers
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.
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.