Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Explanation:
The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.
Total pressure = 2atm
Mole Fraction = number of moles / total number of moles
Neon
Mole Fraction = 4.46 / 7.35 = 0.607
Partial Pressure = 0.607 * 2 = 1.214 atm
Argon
Mole Fraction = 0.74 / 7.35 = 0.101
Partial Pressure = 0.101 * 2 = 0.202 atm
Xenon
Mole Fraction = 2.15 / 7.35 = 0.293
Partial Pressure = 0.293 * 2 = 0.586 atm
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