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A mixture of two gases exerts a total pressure of 3.65 atm. If the partial pressure of one of the gases is 1.86 atm, what is the partial pressure of the other gas in the mixture?

Answer
0.510 atm
1.79 atm
1.96 atm
6.79 atm


If 45 g of LiF are dissolved in 1.8 kg of water, what would be the expected change in boiling point? The boiling point constant for water (Kb) is 0.51 °C/m.
Answer
0.49°C
0.98°C
1.9°C
3.5°C

User Jeff Hykin
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2 Answers

2 votes

Answer: 1. 1.79 atm

2.
0.98^0C

Step-by-step explanation:

1. According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.


p_(total)=p_A+p_B

Given:
p_(total)=3.65atm


p_(A)=1.86atm


p_(B)= ?atm

Thus
3.65=1.86atm+p_(B)


p_(B)=1.79atm

Thus the partial pressure of the other gas in the mixture is 1.79 atm

2.

Elevation in boiling point is:


\Delta T_b=i* k_b* m


\Delta T_b=i* k_b* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}


\Delta T_b = elevation in boiling point

i = Van'T Hoff factor


k_b = boiling point constant

m = molality

For
LiF, i= 2 as it dissociates to give two ions.


\Delta T_b=2* 0.51* (45g)/( 26g/mol* 1.8kg)


\Delta T_b=2* 0.51* (45g)/(26g/mol* 1.8kg)


\Delta T_b=0.98^0C

The expected change in boiling point is
0.98^0C

User Jvans
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7.1k points
6 votes
A mixture of two gases exerts a total pressure of 3.65 atm. If the partial pressure of one of the gases is 1.86 atm, what is the partial pressure of the other gas in the mixture?
Answer

3.65 - 1.86 = 1.79 atm



If 45 g of LiF are dissolved in 1.8 kg of water, what would be the expected change in boiling point? The boiling point constant for water (Kb) is 0.51 °C/m.
Answer

ΔTb = kb(m) = 0.51 ((45/25.94) / 1.8 ) = 0.49 °C

User Extrabacon
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7.0k points