Difference in solubility products and molar solubility?

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

Structure of isobutyl formate

Aliun [14]

Answer:

Isobutyl formate (2-methylpropyl methanoate) is an organic ester with the chemical formula C5H10O2. Used for flavor and fragrance because of its odor.

Explanation:

4 0

3 years ago

Which of these statements are true? Select all that apply.

777dan777 [17]

Answer:

The AH+ for Br2() is 0 kJ/mol.

The AH+ for Hg() is 0 kJ/mol.

The AH; for N261) is 0 kJ/mol.

Explanation:

7 0

2 years ago

J.j thomson a british physicist was the first to identify the

Shalnov [3]

J.j thomson a british physicist was the first to identify the electron in 1987

5 0

3 years ago

A gas has a pressure of 1.34 atm when the temperature is 237K. The gas is then heated until the temperature measures 312K. What

alexandr1967 [171]

The answer for the following question is answered below.

Therefore the new pressure of the gas is 1.76 atm.

Explanation:

Given:

Initial pressure of the gas = 1.34 atm

Initial temperature of the gas = 273 K

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

P ∝ T

$\frac{P}{T}$ = 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

Therefore the new pressure of the gas is 1.76 atm.

6 0

3 years ago