How many moles of a nonvolatile, nonelectrolyte solute are required to lower the freezing point of 1000 grams of water by 5.58

Explanation:

Using freezing point depression formula,

ΔTemp.f = Kf * b * i

Where,

ΔTemp.f = temp.f(pure solvent) - temp.f(solution)

b = molality

i = van't Hoff factor

Kf = cryoscopic constant

= 1.86°C/m for water

= (0 - (-5.58))/1.86

= 3.00 mol/kg

Assume 1 kg of water(solvent)

= (3.00 x 1)

= 3.00 mol.

the correct answer to this question is 2