0.636 g of acetic acid (molar mass = 60 g/mol) is dissolved in 1 kg of water, the solution freezes at $-0.0205^\circ$C.
Calculate van't Hoff factor of acetic acid ($\text{K}_f$ water = 1.86 K-kg/mol)

Given:
Mass of acetic acid, $w = 0.636$ g
Molar mass of acetic acid, $M = 60$ g/mol
Mass of solvent (water), $W = 1$ kg
Depression in freezing point, $\Delta T_f = 0 - (-0.0205) = 0.0205$ K
Cryoscopic constant, $K_f = 1.86$ K kg/mol

The molality ($m$) of the solution is calculated as:
$m = \dfrac{w}{M \times W} = \dfrac{0.636}{60 \times 1} = 0.0106$ mol/kg

Using the formula for depression in freezing point:
$\Delta T_f = i \times K_f \times m$

Substituting the given values:
$0.0205 = i \times 1.86 \times 0.0106$

$0.0205 = i \times 0.019716$

$i = \dfrac{0.0205}{0.019716} \approx 1.041$

Answer: $1.041$