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3 years ago

RAOULT'S LAW

Chemistry

2 answers:

Ugo [173]3 years ago

8 0

Answer:

23.4 torr

Explanation:

For solutions that contain non-volatile solutes, the vapor pressure of the solution can be determined by using the mole fraction of the solvent and the vapor pressure of the pure solvent at the same temperature.

sol

solvent

⋅

∘

solvent

, where

sol

is the vapor pressure of the solution

solvent

is the mole fraction of the solvent

∘

solvent

is the vapor pressure of the pure solvent

In your case, you know that the vapor pressure of pure water at

∘

is equal to

23.8

torr. This means that all you have to do is determine the mole fraction of water in the solution.

As you know, mole fraction is defined as the number of moles of a component of a solution divided by the total number of moles present in that solution.

Use glucose and water's respective molar masses to determine how many moles of each you have

18.0

⋅

1 mole glucose

180.0

0.100 moles glucose

and

95.0

⋅

1 mole water

18.015

5.273 moles water

The total number of moles present in the solution will be

total

glucose

water

total

0.100

5.273

5.373 moles

This means that the mole fraction of water will be

water

5.273

moles

5.373

moles

0.9814

Finally, the vapor pressure of the solution will be

sol

0.9814

⋅

23.8 torr

23.4 torr

The answer is rounded to three sig figs.

Artyom0805 [142]3 years ago

7 0

K so the charcter is 121.3

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Paul [167]

Answer:

The value of Kc is 660 (Option E) is correct

Explanation:

Step 1: Data given

Kp = 27

Temperature = 25.0 °C

Step 2: The balanced equation

2 NO(g) + Br2(g) ⇄ 2 NOBr(g)

Step 3: Calculate Kc

Kp = Kc * (RT)^Δn

⇒ with Kp = 27

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = The temperature = 25 °C = 298 K

⇒ Δn = the difference in moles = -1

27 = Kc * (0.08206*298)^-1

Kc = 660

The value of Kc is 660 (Option E) is correct

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Tom [10]

Answer: The amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

Explanation:

The chemical equation for the reaction of $B_5H_9$ with oxygen gas follows:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ$

To calculate the amount of heat released for the given amount of $B_5H_9(l)$ , we use unitary method, we get:

When 2 moles of $B_5H_9(l)$ reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of $B_5H_9(l)$ will react, the amount of heat released will be = $\frac{5259}{2}\times 0.211=554.8kJ$

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