What is the freezing point of a .743 m aqueous solution of KCl
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Answer:- 14.9 M
Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.
Density of the solution is given as 0.90 grams per mL.
From the mass and density we could calculate the volume of the solution as:
= 111 mL
Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.
= 0.111 L
Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.
= 1.65 mole
To calculate the molarity we divide the moles of ammonia by the liters of solution:
= 14.9 M
So, the molarity of the given commercial sample of ammonia is 14.9 M.