Each axis on a graph should be:<br> equal<br> O numbered<br> Olabeled<br> vertical

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

The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction Reduction potential (V) Li+(aq)+e−→

kakasveta [241]

Under standard conditions :

E(cell) = E(cathode) - E(anode)

Note : cathode has the larger numeric value and anode has the smaller. Therefore

E(cell) = +1.36V - ( -3.04V)

= 1.36 + 3.04

= +4.40V

6 0

3 years ago

Rank the following groups in order of decreasing priority: A. bondCtbondCH B. bondCH(CH3)2 C. bondCH2CH3 D. bondCHdbondCH2

Lelu [443]

Answer:

(CH2)3CH3 > CH2CH2CH3 > CH2CH3 > CH3

Explanation:

Giving the following ; CH3, CH2CH3, CH2CH2CH3, (CH2)3CH3

Priority increases as the number of CH2 group increases and vice versa, as such the one with more CH2 group will be the highest priority and the least compound with the small CH2 group attached, will have the smallest priority.

The arrangement is as follows ; (CH2)3CH3 > CH2CH2CH3 > CH2CH3 > CH3

7 0

3 years ago

20)

anyanavicka [17]

Answer:

D

Explanation:

7 0

3 years ago

Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?

LiRa [457]

Answer:

True

Explanation:

The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

5 0

3 years ago

Each axis on a graph should be: equal O numbered Olabeled vertical