Which of the following diatomic gases has the shortest bond between its two atoms?

A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

5 0

3 years ago

Silicon has three naturally occurring isotopes. Silicon-28 has a mass of 27.98 and a relative abundance of 92.23%. Silicon-29 ha

Advocard [28]

The weighted average atomic mass of Silicon is 28.1

(27.98 x .9223) + (28.98 x .0468) + (29.97 x .0309)
= 25.81 + 1.36 + .93
= 28.1

5 0

3 years ago

Read 2 more answers

What are the Physical changes for oxygen?

wolverine [178]

Answer:

Oxygen is a colorless, odorless, tasteless gas. It changes from a gas to a liquid at a temperature of -182.96°C (-297.33°F). The liquid formed has a slightly bluish color to it. Liquid oxygen can then be solidified or frozen at a temperature of -218.4°C (-361.2°F).

4 0

2 years ago

Explain why the several different types of microscopes are all necessary.

givi [52]

The different types of microscopes are all necessary because not all experiments require the same level of magnification. For dissections low magnification is sufficient, so a dissecting microscope works very well, while for viewing single cells the 1000 fold magnification of a compound light microscope is more accurate.

4 0

3 years ago

How many grams are in 5.2 moles of Li2SO4

skelet666 [1.2K]

Answer:

572 g

Explanation:

Molar mass is the mass of 1 mol of an element or compound

molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound

molar masses of each element making up lithium sulphate

Li - 7 g/mol

S - 32 g/mol

O - 16 g/mol

molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )

molar mass = 110 g/mol

mass of 1 mol of Li₂SO₄ is 110 g

therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g

mass is 572 g

7 0

3 years ago