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Marysya12 [62]
3 years ago
10

Suppose that you add 28.8 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a

dded solute, you find that there is a freezing point depression of 3.06 oC compared to pure benzene. What is the molar mass of the unknown compound
Chemistry
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

THE MOLAR MASS OF THE UNKNOWN COMPOUND IS 242.02 g/mol.

Explanation:

First:

Calculate the change in freezing point:

          Freezing point of pure benzene = 5.5°C

Change in temperature = 5.5 - 3.06 = 2.44 °C

Second:

Using the formula:

Δt = i Kf m

Let's assume i = 1

Kf = 5.12 °C/m

M = x / 0.250 kg of benzene

Then we can calculate x which is the molarity

Re-arranging the formula, we have:

m = Δt / i Kf

x / 0.250 = 2.44 / 1 * 5.12

x = 2.44 * 0.250 / 5.12

x = 0.61 / 5.12

x = 0.119 M

Since it is well known that molarity is the mass of a substance divided by its molar mass. We can then calculate the molar mass.

Molar mass = Mass / molarity

Molar mass = 28.8 g / 0.119 M

Molar mass =242.02 g/mol

Hence, the molar mass of the unknown molecular compound is 242.02 g/mol.

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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
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Answer:

The answer to your question is:

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b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

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                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

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b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

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                        64 g of O2 ------------------  44 g of CO2

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