125 mL of 25 dergree celsius of distilled water was used to wash and transfer a precipitate. rather than 20 ml of CHILLED distilled water (at 4 degree celcius).
the solubility of baruim iodate monohydrate in 25 degree celcius water is 0.028g per 100 ml of water; in 4 degree celcius water, it is 0.010g per 100 ml of water.
(1) what mass of product would you expect to isolate?
(2) calculate the percent error as a result of using 125 ml of 25 degree celsius water, compared with the correct yield using 20 ml of 4 degree celcius water
Actually this is what follow a question that was already posted by someone names "Aymal"
CAN U PLEASE HELP
I gave "Aymal" the instructions for working the problem.
If you work the problem, you will determine the theoretical amount of Ba(IO3)2 that will ppt. I didn't work the problem and come up with an answer so let's call the theoretical amount, TA.
The amount you lose by washing with 125 mL is [0.028 x 125/100] = xx grams
So the amount of Ba(IO3)2 you isolate will be TA-xx.
The second part is a little confusing. You don't get the "correct yield" by using 20 mL of 4 oC because it is somewhat soluble in that water also, although the amount may be negligible if you work through the numbers. But if we assume that the lost amount is as stated above, then
%error = (amount recovered/TA)*100 = ??
By the way, note the correct spelling of celsius.
Certainly! I can help you with that. Let's break down the problem step by step.
(1) To calculate the mass of the product that you would expect to isolate, we need to use the given solubility values and volumes of water used.
The solubility of barium iodate monohydrate in 25°C water is 0.028g per 100 mL of water. Since you used 125 mL of water, the amount of barium iodate monohydrate that can dissolve in this amount of water is:
0.028 g/100 mL * 125 mL = 0.035 g
Therefore, with 125 mL of 25°C water, you would expect to isolate 0.035 g of the product.
(2) To calculate the percent error as a result of using 125 mL of 25°C water instead of 20 mL of 4°C water, we need to compare the expected yield with the correct yield.
The correct yield can be calculated using the solubility of barium iodate monohydrate in 4°C water, which is 0.010 g per 100 mL of water. Since you used 20 mL of water, the amount of barium iodate monohydrate that can dissolve in this amount of water is:
0.010 g/100 mL * 20 mL = 0.002 g
The percent error can then be calculated using the formula:
% Error = (|Expected Yield - Correct Yield| / Correct Yield) * 100
% Error = (|0.035 g - 0.002 g| / 0.002 g) * 100
% Error = (0.033 g / 0.002 g) * 100
% Error = 1650%
So, using 125 mL of 25°C water instead of 20 mL of 4°C water results in a percent error of 1650%.
I hope this helps! Let me know if you have any further questions.