Chemistry - Liquid Solution Question with Solution | TestHub

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:
$m=\frac{w}{M\times W}=\frac{0.636}{60\times 1}=0.0106$ mol/kg

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

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

$0.0205=i\times 0.019716$

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

Answer: $1.041$