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

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

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kobenhavn kobenhavn · Answer 2 · 2019-06-18T06:00:23-04:00

Answer: 1. 1.79 atm

2. \(0.98^0C\)

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:

\(\Delta T_b=i\times k_b\times m\)

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

\(\Delta T_b\) = 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.

\(\Delta T_b=2\times 0.51\times \frac{45g}{ 26g/mol\times 1.8kg}\)

\(\Delta T_b=2\times 0.51\times \frac{45g}{26g/mol\times 1.8kg}\)

\(\Delta T_b=0.98^0C\)

The expected change in boiling point is \(0.98^0C\)