A solution containing 0.1 mol of naphthalene and 0.9 mol of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto `353 K` where its vapour pressure was found to be `670 mm`. The freezing point and boiling point of benzene are `278.5 K` and `353 K` respectively, and its enthalpy of fusion is `10.67 KJ "mol"^(-1)`. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behaviour.

Explanation

To solve the problem step-by-step, we will follow the reasoning and calculations as outlined in the video transcript.

### Step 1: Identify the Given Data
- Moles of naphthalene (N2) = 0.1 mol
- Moles of benzene (N1) = 0.9 mol
- Vapour pressure of pure benzene (P0) = 760 mm
- Vapour pressure of the solution (Ps) = 670 mm
- Freezing point of benzene (Tf0) = 278.5 K
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