Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
Answer: 24.1%, under below assumptions.
Justification:
The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:
- sodium bicarbonate, and
- ammonium bicarbonate
.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.
Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.
With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.
In formulas:
100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%
=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate
=> % sodium bicarbonate = 100% - 75.9% = 24.1%
Answer: 24.1%
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Move more freely :) hope this helps!
Answer:
i don’t understand what you are saying?
Explanation:
