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3 years ago

Three of the primary components of air are

Chemistry

1 answer:

iris [78.8K]3 years ago

5 0

Answer:

174.402 Torr is the partial pressure of oxygen gas.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

$P_T=p_{1}+p_{2}+p_{3}...p_{n}$

where,

$P_T$ = Total pressure

$p_{1}$ = partial pressure of gas-1

$p_{2}$ = partial pressure of gas-2

$p_{3}$ = partial pressure of gas-3

$p_{n}$ = partial pressure of nth gas in the mixture

We have:

$P_T= 1 atm = 760 Torr$

$p_{CO_2}=0.285 Torr$

$p_{N_2}=585.313 Torr$

$p_{O_2}=?$

Now put all the given values is expression, we get the partial pressure of the $O_2$ gas.

$P_T=p_{CO_2}+p_{N_2}+p_{O_2}$

$760 Torr=0.285 Torr+585.313 Torr+p_{O_2}$

$p_{O_2}=174.402 Torr$

174.402 Torr is the partial pressure of oxygen gas.

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3 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$