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The vapour pressure of pure benzene at ...

The vapour pressure of pure benzene at `25^@C` is 640.0 mm Hg and vapour pressure of a solution of a solute in benzene is `25^@C` is 632.0 mm Hg. Find the freezing point of the solution if `k_f` for benzene is 5.12 K/m. `(T_("benzene")^@ = 5.5^@C)`

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To solve the problem step by step, we will follow the necessary calculations to find the freezing point of the solution. ### Step 1: Calculate the Relative Lowering of Vapor Pressure The relative lowering of vapor pressure can be calculated using the formula: \[ \text{Relative Lowering} = \frac{P_0 - P_s}{P_0} \] ...
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The vapour pressure of pure benzene is 639.7 mm Hg and the vapour pressure of solution of a solute in benzene at the temperature is 631.9 mm Hg . Calculate the molality of the solution.

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Knowledge Check

  • The vapour pressure of benzene at 25^@C is 63.7 mm of Hg and the vapour pressure of a solution of a solute in C_6H_6 at the same temperature is 63.9 mm of Hg. The molality of solution is

    A
    0.269
    B
    0.158
    C
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    D
    0.108
  • The vapour pressure of pure benzene at 25^@C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution of

    A
    0.2 m
    B
    0.4 m
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    0.5 m
    D
    0.1 m
  • Benzene and toluene from an ideal solution. The vapour pressure of benzene at 55^(@) C is 400 mm Hg while the vapour pressure of toluene at 55^(@)C is 130 mm Hg. What is the vapour pressure of a solution consisiting of 0.5 mole fraction of benzene and 0.5 mole fraction of toluene at 55^(@)C ?

    A
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    C
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