What is the expected freezing-point depression for a solution that contains 2.0 mol of KCI
1 answer:
Answer:
a. -7.44 °C
Explanation:
Hello there!
In this case, since the freezing point depression formula is:
Thus, since the Van't Hoff's factor is 2 for KCl as it ionizes in K⁺ and Cl⁻, the molarity is 2.0 m (2.0mol/1.0kg) and the freezing point depression constant is 1.86 °C/m, we calculate the freezing point depression as follows:
Therefore, the answer is a. -7.44 °C.
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Explanation:
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Explanation:
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In each mole of carbon dioxide there will be one mole of O₂.
Let us calculate the moles of carbon dioxide gas present first
The conditions are NTP it means , Temperature = 293 K and P = 1 atm
We will use ideal gas equation
PV= nRT
Where
P = Pressure of gas = 1 atm
V= 112mL=0.112L
R= gas constant =0.0821 L atm /mol K
n = moles = ?
Putting values
moles = 0.00466
Thus moles of carbon dioxide will be 0.00466
The moles of O₂ = 0.00466