Question

calculate the depression of freezing point of water when 10g CH3COOH is dissolved in 250g ( Ka= 1.4×10^-3 K 1=1.86 k kg mol^-1).

Answer 1

Depression of freezing point $=K_f\times m$

where m is the molality

let's calculate the molality here which is moles of solute contained in 1000 g of solvent so here acetic acid being the solute and water being the solvent

m=$\dfrac{\frac{10}{60}}{\frac{250}{1000}}$ =0.67

$\Delta T_f=$ 1.86 x 0.67 = 1.25