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Answer : The volume of oxygen occupy at 173° would be, 703.2 mL
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:

Now put all the given values in above equation, we get:

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL
The answer is 4.9 moles.
Solution:
Using the equation for boiling point elevation Δt,
Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,
m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
m = number of moles / 1.0kg
number of moles = m*1.0kg
= (4.883 mol/kg) * (1.0kg)
= 4.9 moles
Answer:
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