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Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
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