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torisob [31]
3 years ago
12

a solution of unknown molecular substance is prepared by dissolving 0.50g of the unknown in 8.0g of benzene. the solution freeze

s at 3.9. determine the molar mass of the unknown
Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.

Explanation:

Mass of the unknown substance = 0.50 g

Freezing point of the solution = 3.9 °C

Freezing point of pure benzene = 5.5 °C

Freezing point dissociation constant Kf = 5.12°C/m

First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.

Change in temperature = 5.5 -3.9 = 1.6 °C

Next is to calculate the number of moles or molarity of the compound that dissolved.

Using the formula:

Δt = i Kf m

Assume i = 1

So,

1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene

x = 1.6 * 0.008 / 5.12

x = 0.0128 / 5.12

x = 0.0025 moles.

Next is to calculate the molar mass using the formula, molarity = mass / molar mass

Molar mass = mass / molarity

Molar mass = 0.50 g /0.0025

Molar mass = 200 g/mol

Hence, the molar mass of the unknown compound is 200 g/mol

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The answer for the following question is answered below.

  • <em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

Explanation:

Given:

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final temperature of the gas = 312 K

To solve:

Final temperature of the gas

We know;

From the ideal gas equation

P × V = n × R × T

So;

from the above equation we can say that

    <em>P ∝ T</em>

     \frac{P}{T} = constant  

     \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1} = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of  a gas

    P_{2} = \frac{1.34*312}{237}

    P_{2}  = 1.76 atm

<em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

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Answer:

See explanation

Explanation:

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B. The equations that turn sulfur to sulfuric acid are;

S(l) + O2(g) -----> SO2(g)

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SO3(g) + H2SO4(l) -----> H2S2O7(l)

H2S2O7(l) + H2O(l) ----> 2H2SO4(l)

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At pH 6, H^+ concentration = 1 * 10^-6 M

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Hence;

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