How many moles of oxygen are prosent in 169 of oxygen gas?
2 answers:
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Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or )
Given : kf = 1.86
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
Data:
p (pressure) = 81.8 kPa = 81.8*10³ Pa ≈ 8.07 atm
v (volume) = ? (in L)
n (number of mols) = 0.352 mol
R (Gas constant) = 0.082 (atm*L/mol*K)
T (temperature) = 25ºC converting to Kelvin, we have:
TK = TC + 273 → TK = 25 + 273 → TK = 298
Formula:
Solving:
p (pressure) = 81.8 kPa = 81.8*10³ Pa ≈ 8.07 atm
v (volume) = ? (in L)
n (number of mols) = 0.352 mol
R (Gas constant) = 0.082 (atm*L/mol*K)
T (temperature) = 25ºC converting to Kelvin, we have:
TK = TC + 273 → TK = 25 + 273 → TK = 298
Formula:
Solving: