on heating 25g of a saturated solution to dryness at 60c,4g of anhydrous salt was recovered.calculate it's solubility in grammes per 100g of solvent
Solute = 4g
Solvent = 25 - 4 = 21g
Solubility in 100g of solvent;
21g = 4g
100g = x
; x = 100 × 4/21
= 400/21
= 19.05 g/100g of solvent
Mass of saturated solution= 25g
Mass of solute=4g
Mass of solvent=25g-4g
=21g
Density=mass/volume
Note: Density of Water =1g/cm3
Volume= Mass/Density
= 21/1
= 21cm3
Solubility (g/dm3)= mass of solute/(volume of solvent/1000)
But since the question asked for the solubility in gramme per 100g of the solvent, therefore the volume of the solvent would be divided by 100 not 1000
Solubility= 4/(21/100)
= 4/0.21
= 19.0476
= 19.05g/100g of solvent
Obviously right
Yes, that's correct! Your calculation for the solubility of the salt in grams per 100 grams of solvent is accurate. It is equal to 19.05 g/100g of solvent.
I'm sorry, I don't understand what you're trying to say. Can you please provide more context or rephrase your statement?
What is ai
To calculate the solubility in grams per 100g of solvent, you need to know the amount of solute (in this case, anhydrous salt) and the amount of solvent used.
In this scenario, you started with a saturated solution containing 25g of the solute. By heating it to dryness at 60°C, you removed the water (solvent) and left behind 4g of anhydrous salt.
To calculate the solubility, divide the amount of solute by the amount of solvent used. In this case, the solvent used is the amount of water that was initially in the saturated solution.
To determine the amount of solvent in the saturated solution, subtract the recovered anhydrous salt (4g) from the initial solute (25g):
Solvent used = Initial solute - Recovered anhydrous salt
= 25g - 4g
= 21g
Now you can calculate the solubility of the anhydrous salt in grams per 100g of solvent:
Solubility = (Recovered anhydrous salt / Solvent used) x 100
= (4g / 21g) x 100
≈ 19.05 grams per 100 grams of solvent
Therefore, the solubility of the anhydrous salt is approximately 19.05 grams per 100 grams of solvent.