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4 years ago

If a 0.710 m 0.710 m aqueous solution freezes at − 2.00 ∘ C, −2.00 ∘C, what is the van't Hoff factor, i , i, of the solute?

Chemistry

1 answer:

Lena [83]4 years ago

4 0

Answer:

The van't Hoff factor of the solute is 1.51

Explanation:

Step 1: Data given

Molality = 0.710 molal

The aqueous solution freezes at − 2.00°C

Freezing point depression constant of water = 1.86 °C/m

Step 2: Calculate the van't Hoff factor

ΔT = i*Kf * m

⇒ with ΔT = The difference between the feezing point of pure and solution = 2.00°C

⇒ i the van't Hoff factor = TO BE DETERMINED

⇒ Kf = Freezing point depression constant of water = 1.86 °C/m

⇒ m = the molality of the solution = 0.710 molal

2.00 = i * 1.86 * 0.710

i = 1.51

The van't Hoff factor of the solute is 1.51

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3 years ago

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p

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1) to calculate the limiting reactant you need to pass grams to moles.
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mass of N2O4: 50.0 g
molar mass of N2O4 = 92.02 g/mol
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mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
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By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

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There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
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1,62 mol of N2= mass/ 28
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4 0

3 years ago

in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125

PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

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