ACS Gen Chem Questions

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xjoohn

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Can someone answer and explain to me these two questions? Thanks...

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21. Basically you are making a saturated solution, then cooling it which decreases the solubility and solid potassium chromate will precipitate. So, look at your graph at 60 degrees C, the solubility is 40 g K2Cr2O7 / 100 mg water, since you started with 100 g K2Cr2O7 and 200 g water, you actually have 80 g in 200 g water and 20 grams left over as solid in bottom of beaker. Then it says the solution is poured off (decanted) into a new beaker so you have a saturated solution of 80 g K2Cr2O7 in 200 g water, and you left the 20 g solid in the original beaker. Now you cool that solution down to 20C. Again, look at solubility on graph which is about 7.5g/100 g water, therefore with 200 g water that means your solution contains about 15 g of K2Cr2O7 and 25 g should precipitate (40g @40C minus 15g @20C =25 g precipitate. or 24 g which is A.
 
11. Basically, this is the definition of "normal boiling point" since Boiling occurs where the vapor pressure of the gas = atmospheric pressure (760 mmHg = 760 torr = 1 atm) Therefore IF you are precisely at sea level, and atmospheric pressure is exactly 1 atm the boiling point would appear to be 78C and the graph appears accurate. However, since you do not KNOW that the liquid is at 1 atm, and you cannot prove this, I'd say B. Statement probably true but additional data needed for final decision.

What you need to know is the definition of boiling point in terms of vapor pressure and atmospheric pressure.
 
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