Suppose that you add 27.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.69 oC compared to pure benzene. What is the molar mass of the unknown compound

Molar mass is the mass of the one mole of substance. The molar mass of the given unknown compound is 153.3 g/mol.

Molality of the compound can be calculated using

ΔT = i Kf m

Where,

ΔT = freezing point depression = 3.69 °C

i =  Van't Hoff factor of Benzene = 1

Kf =  constant of freezing = 5.12 °C/m

m = molality = ?

Put the values in the equation,

3.69 = 1 x 5.12 x m

m = 0.72 molal

Number of moles of the compound,

So, molar mass of the unknown compound,

The molar mass of the given unknown compound is 153.3 g/mol.

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The molar mass of the unknown compound is 153.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown molecular compound = 27.6 grams

Mass of benzene =  0.250 kg

Kf of benzene = 5.12 °C/m

freezing point depression of 3.69 °C

Step 2:  Calculate molality

ΔT = i*Kf*m

⇒with ΔT = reezing point depression of 3.69 °C

⇒with i = the van't Hoff factor of Benzene = 1

⇒with Kf = 5.12 °C/m

⇒ with m = molality = moles unknown compound / mass of benzene

3.69 = 1 * 5.12 * m

m = 0.72 molal

Step 3: Calculate moles of the unknown compound

molality = moles / mass benzene

0.72 molal = moles / 0.250 kg

Moles = 0.72 m * 0.250 kg

Moles = 0.18 moles

Step 4: Calculate molar mass of the unknown compound

molar mass = mass / moles

Molar mass = 27.6 grams / 0.18 moles

Molar mass = 153.3 g/mol

The molar mass of the unknown compound is 153.3 g/mol

You have to calculate the oxidation estates of the atoms in each compound.

I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.

In K2Cr2O7:

- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.

- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.

That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.

In Cr2O3:

- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6

- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.

So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.

Cr has a change in oxidation number of - 3.