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Charra [1.4K]
3 years ago
8

The normal freezing point of a certain liquidXis0.4°C, but when5.90gof ureaNH22COare dissolved in450.gofX, it is found that the

solution freezes at−0.5°Cinstead. Use this information to calculate the molal freezing point depression constantKfofX.
Chemistry
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

The molal freezing point depression constant for X is 4.12 °C/m

Explanation:

Apply the colligative property of freezing point depression:

ΔT = Kf . m . i

We assume the X lquid is non electrolytic, so i = 1

ΔT = T° freezing point of solvent pure - T° freezing point of solution

m = molality (moles of solute in 1kg of solvent)

Kf = Cryoscopic constant (the data we were asked for)

ΔT = 0.4° - (- 0.5°) = 0.9°

Let's calculate molality

Molar mass of urea = 60.06 g/m

Moles of urea = mass of urea / molar mass

5.90 g / 60.06 g/m = 0.098 moles

This moles are in 450 g of solvent, prepare a rule of three to find out the moles in 1000 g

450 g ____ 0.098 m

1000 g ____ ( 1000.  0.098)/450 = 0.218 m

0.9° = Kf . 0.218 m . 1

0.9° / 0.218 m = Kf → 4.12 °C/m

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This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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The following questions pertain to a system contains 122 g CO(g) in a 0.400 L container at -71.2 degrees C.
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Answer:

a. P = 182 atm

Explanation:

Data Given:

amount of CO = 122g

Volume of CO = .400 L

Temperature of CO =  -71.2°C

Convert the temperature to Kelvin

T = °C + 273

T =  -71.2 + 273

T =  201.8 K

a. Calculate the pressure exerted by the CO(g) in this system using the ideal gas equation (P) = ?

Solution:

To calculate Pressure by using ideal gas formula

                  PV = nRT

Rearrange the equation for Pressure

                   P = nRT / V . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant = 0.08206 L.atm / mol. K

For this we have to know the mole of the gas and the following formula will be used

                 no. of moles = mass in grams / molar mass . . . . . . (2)

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in equation 2

                no. of moles = 122 g / 28 g/mol

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Now put the value in formula (1) to calculate Pressure for CO

P = 4.4 x 201.8 K x 0.08206 (L.atm/mol. K) / 0.400 L

P = 182 atm

So the pressure will be 182 atm

__________

b. Data Given:

Actual pressure exerted by CO = 145 atm

expected pressure exerted by CO = 182 atm

why the actual pressure is less than what would be expected = ?

Explanation:

This is because of the deviation from ideal behavior of real gases.

The real gases approach to ideal behavior under very high temperature and very low pressure.

But CO deviate from ideal behavior to give expected value for pressure, because it behave at high pressure and low temperature.

This non-ideal behavior is due to two postulate of ideal behavior

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  • Gas molecules have negligible inter-molecular interaction

but these postulates not obeyed under real condition. so we calculated the pressure using ideal condition values for gas and obtained the expected value for pressure but the actual pressure value was detected under normal condition.

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Answer:

A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom.

Explanation:

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