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MAXImum [283]
3 years ago
13

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other h

as carbon tetrachloride as the solvent. Determine the molal concentration, m (or b ), and boiling point, T b
Chemistry
1 answer:
Degger [83]3 years ago
6 0

Answer:

Molality of both benzene and carbon tetrachloride is 1.32 m .

boiling point is 83.4396 °C

Explanation:

When a non-volaile solute is added to a solvent, its boiling point increases. The relationship increase in temperature and molality of the solute is given as follows:

ΔT = m × Kb

Boiling point of pure benzene is 80.1 °C.

Molal elevation boiling constant of benzene (k_b) is 2.53 °C/m

Increase in boiling point of benzene = m × 2.53

Therefore, Boiling point of benzene solution = 80.1 + 2.53m

Boiling point of pure carbon tetrachloride is 76.8 °C

Molal elevation boiling constant of carbon tetrachloride (k_b) is 5.03 °C/m

Increase in boiling point of benzene = m × 5.03

Therefore, Boiling point of carbon tetrachloride solution = 76.8 + 5.03m

It is given that boiling point of both the solutions are same therefore,

76.8 + 5.03m = 80.1 + 2.53m

5.03m - 2.53m = 80.1 - 76.8

2.5m = 3.3

m = 1.32  

Molality of both benzene and carbon tetrachloride is 1.32 m

Boiling point of the solutions = 80.1 + 2.53 × 1.32

                                            = 83.4396 °C

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Read 2 more answers
A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0
3 years ago
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