What is the molality of 22.5 g acetone
(CH3)2CO , in 200 g water ?
( the answer is 22.0 m , that I think is wrong .. )
m= moles/kg solvent= (22.5/58)/.200
I get considerably less than 22.
but 22.5/58 = 0.3879 , and 0.3879/0.2 = 1.9
but the answer is another (22) .. I can't get it
The answer in the data base/text/whatever must be wrong. I get 1.93965 which I would round to 1.94 to 3 s.f.
my answer is 1.9 too 😑
To find the molality (m) of a solution, we need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).
First, determine the number of moles of acetone (CH3)2CO. We can use the molar mass of acetone to convert grams to moles.
The molar mass of acetone is:
C: 12.01 g/mol
H: 1.01 g/mol (3 hydrogen atoms in acetone)
O: 16.00 g/mol
Adding these up, we get:
12.01 + (1.01 * 3) + 16.00 = 58.08 g/mol
Now, let's calculate the number of moles of acetone:
moles = mass / molar mass
moles = 22.5 g / 58.08 g/mol
Next, calculate the mass of water in kg:
mass of water = 200 g / 1000 = 0.2 kg
Finally, we can calculate the molality using the formula:
molality (m) = moles of solute / mass of solvent (in kg)
molality = (22.5 g / 58.08 g/mol) / 0.2 kg
By simplifying, we get:
molality = 22.5 / (58.08 * 0.2)
Evaluating the expression gives us:
molality ≈ 19.316 m
Therefore, the molality of the solution is approximately 19.316 m, not 22.0 m as you mentioned.