Given: solubility (g/100g H2O)
temp degrees celcius KBR
20 65
40 80
60 90
80 100
100 110
A solution containing 10.5g of KBr in 10g of H2O at 100 degrees celcius is cooled to 40 degrees celcius.
A)calculate the g/100g H2O of KBr
B)is the solution saturated at 100 degrees celcius?
C) Is the solution saturated at 40 degrees celcius?
D) How much solid is formed during the cooling?
A) To calculate the g/100g H2O of KBr at 100 degrees Celsius, we can use the given solubility data. According to the data, the solubility of KBr at 100 degrees Celsius is 110 g/100g H2O.
B) To determine if the solution is saturated at 100 degrees Celsius, we compare the actual concentration to the maximum solubility at that temperature.
The actual concentration is given as 10.5g of KBr in 10g of H2O. To calculate the g/100g H2O, we divide the amount of KBr by the total weight of the solution (KBr + H2O) and multiply by 100:
g/100g H2O = (10.5g KBr / (10.5g KBr + 10g H2O)) * 100
g/100g H2O = (10.5 / 20.5) * 100
g/100g H2O ≈ 51.22
Since the actual concentration is lower than the maximum solubility at 100 degrees Celsius (110 g/100g H2O), the solution is not saturated at this temperature.
C) To determine if the solution is saturated at 40 degrees Celsius, we use the same approach. According to the data, the solubility of KBr at 40 degrees Celsius is 80 g/100g H2O.
Using the same calculation as before, we can find the g/100g H2O at 40 degrees Celsius:
g/100g H2O = (10.5g KBr / (10.5g KBr + 10g H2O)) * 100
g/100g H2O = (10.5 / 20.5) * 100
g/100g H2O ≈ 51.22
Again, the actual concentration is lower than the maximum solubility at 40 degrees Celsius (80 g/100g H2O). Therefore, the solution is not saturated at this temperature either.
D) To calculate how much solid is formed during cooling from 100 degrees Celsius to 40 degrees Celsius, we compare the solubility at these two temperatures.
At 100 degrees Celsius:
Solubility of KBr = 110 g/100g H2O
At 40 degrees Celsius:
Solubility of KBr = 80 g/100g H2O
The difference in solubility between the two temperatures is 110 g/100g H2O - 80 g/100g H2O = 30 g/100g H2O.
Since the solution is not saturated at 40 degrees Celsius, it means that there is room for more KBr to dissolve. Therefore, no solid is formed during the cooling process.
To answer these questions, we need to use the solubility data provided.
A) To calculate the g/100g H2O of KBr, we need to determine the solubility of KBr at 40 degrees Celsius. We can find this information in the given solubility table. According to the table, the solubility of KBr at 40 degrees Celsius is 80 g/100g H2O.
B) To determine whether the solution is saturated at 100 degrees Celsius, we compare the actual concentration with the solubility at that temperature. The solution contains 10.5g of KBr in 10g of H2O, which corresponds to a concentration of 10.5 g/10 g H2O or 105 g/100g H2O. Comparing this to the solubility at 100 degrees Celsius of 110 g/100g H2O, we can conclude that the solution is unsaturated at 100 degrees Celsius because the actual concentration is lower than the solubility.
C) To determine whether the solution is saturated at 40 degrees Celsius, we follow the same process as in step B. The solubility of KBr at 40 degrees Celsius is 80 g/100g H2O. Since our concentration at 40 degrees Celsius is 10.5 g/10 g H2O or 105 g/100g H2O, which is greater than the solubility at that temperature, we can conclude that the solution is saturated at 40 degrees Celsius.
D) To calculate how much solid is formed during cooling, we need to determine how much of the excess KBr will precipitate out. We have 105 g/100g H2O at 40 degrees Celsius, but the solubility of KBr at 40 degrees Celsius is only 80 g/100g H2O. The difference between these two values (105 - 80 = 25 g/100g H2O) represents the excess KBr that will form solid precipitates when the solution is cooled.
Therefore, the answers to your questions are:
A) The g/100g H2O of KBr at 40 degrees Celsius is 80 g/100g H2O.
B) No, the solution is unsaturated at 100 degrees Celsius.
C) Yes, the solution is saturated at 40 degrees Celsius.
D) 25 grams of solid KBr will precipitate out during cooling.
Note the correct spelling of celsius.
a) Convert 10.5 g KBr/10 g H2O to g KBr/100 g H2O, then compare with your table.
10.5 g KBr x (100 gH2O/10 g H2O) = 105 g KBr/100 g H2O. Your table says you can dissolve 110 g KBr at 100.
b)Look at the numbers.
c) Compare the numbers at 40 C.
d)If you had 110 g at 100 C and you can hold 80 g at 40C then 110-80 must be amount that crystallizes when cooled from 100 to 40.