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balu736 [363]
2 years ago
9

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

into naphthalene
Chemistry
1 answer:
lbvjy [14]2 years ago
3 0

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.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

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:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

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

Learn more: brainly.com/question/2292439

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4 0
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3 H2 (g) + N2 (g) 2 NH3 (g)
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Answer:

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Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

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Pressure = standard = 1 atm

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Chemical equation:

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PV  = nRT

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n = 1.36 atm.L / 22.43 atm.L/mol

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                  3          :            2

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