When 45 g of an unknown compound was dissolved in 500 g of water, the solution has freezing point of `-0.93^(@)C`
(i) What is the molecular weight of compound ? `(K_(f)=1.86)`
(ii) If empirical formula is `CH_(2)O`, what is the molecular formula of compound ?

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Calculate the change in freezing point (ΔTf)

The freezing point of pure water is 0°C. The freezing point of the solution is -0.93°C.

\[
\Delta T_f = T_f(\text{solvent}) - T_f(\text{solution}) = 0 - (-0.93) = 0.93 \, \text{°C}
...