[abs]

i was wondering if anyone could help me with my summer chemistry homework. i have no clue how to start or figure out how to do these two problems. help would be greatly appreciated!

(1) A bottle of wine contains 11.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality.
mass percent
 %
molality
  mol/kg

(2)What volume (in mL) of 0.55 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 15.5 mL of 0.45 M Ba(NO3)2 solution?

Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)

  mL

[sdekivit]

i was wondering if anyone could help me with my summer chemistry homework. i have no clue how to start or figure out how to do these two problems. help would be greatly appreciated!

(1) A bottle of wine contains 11.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality.
mass percent
 %
molality
  mol/kg

(2)What volume (in mL) of 0.55 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 15.5 mL of 0.45 M Ba(NO3)2 solution?

Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)

  mL

also here you won't get a solution to the questions. Here are the hints again:

1) the solvent  is water. Assume we have 100 mL wine, then 11,6 mL is ethanol and 88,4 mL is water. You can then calculate the masses using the density of ethanol and water (0,998 kg / L). Molality is given by:

mol solute / kg solvent.

2) 15,5 mL 0,45 M bariumnitrate is . mmol. This also the amount of mols Ba in the molecule. Stoichiometry is 1:1 so we need . mmol sodium sulphate. What is thus the molarity if you know that molarity is mmol/mL (and thus mol/L) ?

[Borek]

First question is very doubtful:

http://www.chembuddy.com/?left=concentration&right=volume-volume-percentage

1) the solvent  is water. Assume we have 100 mL wine, then 11,6 mL is ethanol and 88,4 mL is water.

Not 11.6 but 11.5, but that's simple mistake. There is much more important problem with this approach that leads to incorrect results.

Mixing 11.6 mL of ethanol with 88.4 mL of water you get 99.0 mL of solution. To obtain 100 mL of final solution you have to use 89.4 mL of water.

Unfortunately, without access to density tables such questions just can't be done properly.

[abs]

what am i doing wrong?

11.5 mL ethanol
89.5 mL water

(.789g/mL) * (11.5 mL ethanol)= 9.0735 grams ethanol

(.0895L) * (.998 kg/l)= .089329 kg = 89.329 grams water

mass % = 9.0735g / 98.4025 *100 = 9.2208023

the program that i submit my work through says this is wrong.

[sdekivit]

First question is very doubtful:

http://www.chembuddy.com/?left=concentration&right=volume-volume-percentageNot 11.6 but 11.5, but that's simple mistake. There is much more important problem with this approach that leads to incorrect results.

Mixing 11.6 mL of ethanol with 88.4 mL of water you get 99.0 mL of solution. To obtain 100 mL of final solution you have to use 89.4 mL of water.

Unfortunately, without access to density tables such questions just can't be done properly.

my answer is: 9,32 % --> maybe the program check significant numbers.

[Borek]

what am i doing wrong?

11.5 mL ethanol
89.5 mL water

Check if the results obtained for 11.5mL and 88.5mL will be correct. The problem is, such questions are often asked with the simplistic approach in mind - and thus they should be never asked, as correct answers are not accepted, while answers accepted are wrong.

I hate such things.