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yKpoI14uk [10]
3 years ago
15

Calculate the freezing point of a solution 1.25 g benzene (C6H6) in 125 g of chloroform (CHCl3).

Chemistry
1 answer:
posledela3 years ago
3 0

Answer:

The freezing point for the solution is -64.09°C

Explanation:

This problem can be solved, by the freezing point depression. This colligative problem shows, that the freezing point of a solution is lower than the freezing point of pure solvent.

ΔT = Kf.  m

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

Kf = Cryscopic constant, for chloroform is 4.68

T°freezing pure solvent = -63.5°C

m is mol/kg of solvent → molality

Let's determine the moles of benzene

1.25 g / 78 g/mol = 0.0160 mol

Let's convert the mass of solvent to kg

125 g . 1kg / 1000 g = 0.125 kg

m = 0.0160 mol / 0.125 kg → 0.128 m

Let's go to the formula to replace the data

-63.5°C - T° freezing solution = 4.68 °C/m . 0.128 m

T° freezing solution = - (4.68 °C/m . 0.128 m + 63.5°C)

T° freezing solution = - 64.09°C

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Answer:

8.37g

Explanation:

Step 1 :

The balanced equation for the reaction. This is given below:

N2 + 2O2 —> 2NO2

Step 2:

Data obtained from the question.

Volume (V) of N2 = 2L

Pressure (P) = 840mmHg

Temperature (T) = 24°C

Number of mole (n) of N2 =?

Step 3:

Conversion to appropriate unit.

For pressure:

760mmHg = 1atm

840mmHg = 840/760 = 1.11 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 24°C

T(K) = 24°C + 273

T(K) = 297K

Step 4:

Determination of the number of mole N2.

The number of mole of N2 can be obtained by using the ideal gas equation as follow:

Volume (V) of N2 = 2L

Pressure (P) = 1.11 atm

Temperature (T) = 297K

Number of mole (n) of N2 =?

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.11 x 2 / 0.082 x 297

n = 0.091 mole

Therefore, the number of mole of N2 that reacted is 0.091 mole

Step 5:

Determination of the mass of NO2 produced from the reaction. This is illustrated below:

N2 + 2O2 —> 2NO2

From the balanced equation above,

1 mole of N2 produced 2 moles of NO2.

Therefore, 0.091 mole of N2 will produce = 0.091 x 2 = 0.182 mole of NO2.

Finally, we will convert 0.182 mole of NO2 to gram as shown below:

Number of mole NO2 = 0.182 mole

Molar mass of NO2 = 14 + (16x2) = 46g/mol

Mass = number of mole x molar mass

Mass of NO2 = 0.182 x 46

Mass of NO2 = 8.37g

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