Saturday
May 18, 2013

# Homework Help: Big review 4 finals (2 posts in 1) check my work

Posted by Kizner (chem) on Wednesday, June 14, 2006 at 7:45am.

A 7.58 gram sample of benzoic acid, C6H5CO2H, is dissolved in 100 mL of benzene. The density of benzene is 0.879 g/mL. What is the molality of the benzoic acid?

Sol: Convert mL to grams benzene.

100 mL * (0.879g benzene/1 mL) = 87.9 g

Molality = # molsolute / # kg solvent

So I convert above answer to kg:
87.9 g * (1 kg/1000 g)=.0879 kg Benzene

M.W. Benzene: 122.12 g

?mol Benzene :
7.58g benzene*(1 mol benzene/122.12g)=0.0621 mol Benzene.

Molality =
.0621 mol/.0879 kg =.706 mol/kg benzene

2. What is the molarity of a Ba(OH)2 solution if 15.48 mL of it are required to react with 25 mL of .303 M HCl solution? Show the balanced reaction in your calculations.

Sol:
2HCl(aq)+ Ba(OH)2(aq)-->BaCl2 + 2H2O(L)

(there are some exceptions when dealing with the formula: m1v1 = m2v2; I know to apply when you deal with the same compound. Here we have two. I'll use formula anyways as I see no way to proceed.

(2*.3030M HCl)*.025L = .0155M2
.0152M = m2 ??

A 7.58 gram sample of benzoic acid, C6H5CO2H, is dissolved in 100 mL of benzene. The density of benzene is 0.879 g/mL. What is the molality of the benzoic acid?

Sol: Convert mL to grams benzene.

100 mL * (0.879g benzene/1 mL) = 87.9 g

Molality = # molsolute / # kg solvent

So I convert above answer to kg:
87.9 g * (1 kg/1000 g)=.0879 kg Benzene

M.W. Benzene: 122.12 g
This should be benzoic acid. That is the solute in the problem. Benzene is the solvent and you have already converted that to kg.
?mol Benzene :
7.58g benzene*(1 mol benzene/122.12g)=0.0621 mol Benzene.
Just convert the word benzene to benzoic acid. You used the molar mass of 122.12 correctly for benzoic acid. I suspect this is just a little confusion between the similarity of the two names.
Molality =
.0621 mol/.0879 kg =.706 mol/kg benzene
I would have written 0.0706 m (note the first 0 in front of the decimal AND the small m. Your answer is correct and there is noting wrong with writing it as you did but the other way avoids confusion that sometimes exists. Good work!
2. What is the molarity of a Ba(OH)2 solution if 15.48 mL of it are required to react with 25 mL of .303 M HCl solution? Show the balanced reaction in your calculations.

Sol:
2HCl(aq)+ Ba(OH)2(aq)-->BaCl2 + 2H2O(L)
Since you have used (aq) for everything on the left, I think you should have written (aq) after the BaCl2, also. You have it for everyting else so I suspect this is just an oversight.
(there are some exceptions when dealing with the formula: m1v1 = m2v2; I know to apply when you deal with the same compound. Here we have two. I'll use formula anyways as I see no way to proceed.
You COULD use a modified formula but that can get confusing. I recommend the following because it follows a logical path; i.e., calculate mols HCl, convert using the equation to mols Ba(OH)2, convert mols Ba(OH)2 to M by M = mols/L.
(2*.3030M HCl)*.025L = .0155M2
.0152M = m2 ??

When the coefficients for the balanced equation are not the same, as in this case, it isn't proper to use the formula m1v1 = m2v2. The best way to proceed is as follows:
mols HCl used in the titration is M x L = 0.303 x 0.025 L = 0.00758 mols HCl.
Now convert mols HCl to mols Ba(OH)2 using the balanced chemical equation. The numbers come from the coefficients in the equation.
0.00758 mols HCl x (1 mol Ba(OH)2/2 mols HCl) = 0.00379 mols Ba(0H)2.
Note that this is the same way you converted g to kg in the first problem; i.e., in this case mols HCl in the numerator of the first term cancel with mols HCl in the denominator of the second term and that leaves units of Ba(OH)2 for the answer.
Now convert mols Ba(OH)2 to M remembering that mols = M x L. Solving for M we have M = mols/L = 0.00379/0.01548 L = 0.245 M. Check my math. Check my work. I may not have rounded exactly. I hope this helps.

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