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
8 answers
You started with 25g solution and recovered 4g of salt. So the solubility is 4g per 25g. That would be 4 x 100/25 = solubility in 100 g.
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
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
Obviously right
Fnndnd
I'm sorry, I don't understand what you're trying to say. Can you please provide more context or rephrase your statement?
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
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
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.
What is ai
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