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Alla [95]
3 years ago
6

When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the freezing point of the resulting solution is 3

.77°C. The freezing point of pure benzene is 5.48°C, and the Kf for benzene is 5.12°C/m. What is the molecular weight of the unknown compound?
Chemistry
1 answer:
Lelu [443]3 years ago
5 0

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

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7 0
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How many total atoms are in 0.680 g of P₂O5?
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Answer: 2.88×10^{24}atoms P_{2}O_{5}

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1. For the first step, we should also look at the periodic table to find the molar mass of the compound, then use that as the denominator.

0.680g P_{2} O_{5} *\frac{1molP_{2} O_{5} }{141.948gP_{2} O_{5} } =4.79mol P_{2} O_{5}

2. Now that it is converted to moles, we must convert it to atoms by multiplying it by Avogadro's number.

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With this information, we know that there are 2.88X10^{24} total atoms in 0.680 grams P_{2} O_{5}.

I hope this helps! Pls give brainliest!! :)

8 0
2 years ago
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