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Semenov [28]
2 years ago
13

A solution contains 3.95 g of carbon disulfide (CS2, molar mass = 76.13 g/mol) and 2.43 g of acetone ((CH3)2CO, molar mass = 58.

08 g/mol). The vapor pressure of pure carbon disulfide and acetone at 35 o C are 515 torr and 332 torr, respectively. Assuming ideal solution behavior, calculate the vapor pressure of each of the components and the total vapor pressure above the solution. The total vapor pressure above the solution at 35 o C was experimentally determined to be 645 torr. Is the solution ideal? If not, indicate whether the solution deviates from Raoult’s law in a positive or negative manner
Chemistry
1 answer:
Feliz [49]2 years ago
6 0

Answer:

The solution is not ideal and shows a positive deviation from Raoult’s law since Psolution (experimental) > Psolution (actual).

Explanation:

Number of moles of CS2 = 3.95g/76.13gmol-1 = 0.0519 moles

Number of moles of acetone = 2.43g/58.08gmol-1 = 0.0418 moles

Total number of moles = 0.0937 moles

Mole fraction of CS2 = 0.0519/0.0937 = 0.5538

Mole fraction of acetone = 0.0418/0.0937 = 0.4461

From Raoult’s law;

PCS2 = 0.5538 × 515 torr = 285.207 torr

Pacetone = 0.4461 × 332 torr = 148.1052 torr

Total pressure = 285.207 torr + 148.1052 torr = 433.3 torr

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You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
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Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

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mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

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amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

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