The solubility of LiCl is 83.5 g/100 g H2O at 20 oC , and is 89.8 g / 100 g H2O at 40 oC , then what is the required mass (in gram ) of LiCl to obtain a saturated solution in 300 g H2O at 25 oC ? please explain
thank a lot
Please
Assuming you meant 25C and not 20C or 40C(for which data is available), you must draw a graph with these two points on it then read the solubility at 25 C.
Since the solubility varies only slightly between the two temperatures listed (at 20 and 40) you can come very close calculating the solubility at 25 this way.
83.5 g/100 g H2O at 20C
89.8 g/100 g H2O at 40C
Difference is 89.8-83.5 = 6.3g/100 for a 20C difference (that's 40C-20C). So we can estimate that the solubility for a 5 C difference is
[83.5 + (6.3 g x 5/20)] = about 83.5+1.6 = 85.1 g LiCl/100 g H2O.
Since you want a saturated solution in 300 g H2O that will obviously take 3*85.1 = ? g LiCl at 25C.
To determine the required mass of LiCl to obtain a saturated solution in 300g of water at 25°C, we need to find out how much LiCl can dissolve in 100g of water at 25°C, and then calculate it for 300g of water.
First, let's find the solubility of LiCl at 25°C. We know the solubility at 20°C is 83.5g/100g H2O, and at 40°C is 89.8g/100g H2O. Since our target temperature is between these two, we can assume that the solubility varies linearly with temperature.
To find the solubility at 25°C, we can use linear interpolation.
Step 1: Determine the temperature difference
Temperature difference = (25 - 20) = 5°C
Step 2: Determine the solubility difference
Solubility difference = (89.8 - 83.5) = 6.3g/100g H2O
Step 3: Calculate the solubility at 25°C
Solubility at 25°C = Solubility at 20°C + (Temperature difference × Solubility difference / Temperature difference)
Solubility at 25°C = 83.5g/100g H2O + (5°C × 6.3g/100g H2O / 5°C)
Solubility at 25°C = 83.5g/100g H2O + 6.3g/100g H2O
Solubility at 25°C = 89.8g/100g H2O
So, the solubility of LiCl at 25°C is 89.8g/100g H2O.
Next, we need to calculate the required mass of LiCl to obtain a saturated solution in 300g of water at 25°C.
Step 1: Calculate the mass of LiCl in a saturated solution in 100g of water at 25°C
Mass of LiCl in 100g of water = Solubility at 25°C = 89.8g
Mass of LiCl in 300g of water = 3 × Mass of LiCl in 100g of water
Mass of LiCl in 300g of water = 3 × 89.8g
Mass of LiCl in 300g of water = 269.4g
Therefore, the required mass of LiCl to obtain a saturated solution in 300g of water at 25°C is 269.4 grams.