1)A solid mixture consists of 25.5 g of KNO3 (potassium nitrate) and 4.5 g of K2
ID: 1017692 • Letter: 1
Question
1)A solid mixture consists of 25.5 g of KNO3 (potassium nitrate) and 4.5 g of K2SO4 (potassium sulfate). The mixture is added to 130. g of water.
If the solution described in the introduction is cooled to 0 C what mass of K2SO4 will crystallize?
2) As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.57 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood? [Use the value of the Henry's law constant k calculated in Part A, 6.26×104mol/(Latm).]
Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.
Explanation / Answer
solubility of KNO3 at 0C0 = 14g of KNO3 in 100g of water
Xg of KNO3/130 of water = 14g of KNO3/100g of water
x/130 = 14/100
x = 14*130/100 = 18.2g
if 25.5g of KNO3 in 130g of water at 0C0 and only soluble in 18.2g of KNO3
undissolved KNO3 = 25.5-18.2 = 7.3g of KNO3 is undissolved at 0C0
2. C = KH*P
KH = 6.26×104mol/(Latm).
P = 2.57atm
C = KH*P
= 6.26*10-4 *2.57
= 1.608*10-3 mole/L .>>>>> answer
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