At –10.5 °c (a common temperature for household freezers), what is the maximum mass of fructose (c6h12o6) you can add to 1.00 kg of pure water and still have the solution freeze? assume that fructose is a molecular solid and does not ionize when it dissolves in water. kf values are given here.

For this problem, we use the freezing point depression formula:

Tf,solvent - Tf,solution = Kf×m
Where
Tf,solvent is freezing pt of the solvent
Tf,solution is freezing pt of solution
Kf is the molal freezing point depression constant of the solvent
m is the molality equal to mol solute/kg solvent

For water as solvent, Tf,solvent = 0°C; Kf = 1.86 °C/m:

0°C - ⁻10.5 °C = (1.86 °C/m)(m)
Solving for m,
m = 5.645 mol solute/kg solvent

5.645 = mol solute/1 kg water
mol solute = 5.645 mol fructose
Since the molar massof fructose is 180.16 g/mol,
Mass of Fructose = 5.645 mol * 180.16 g/mol
Mass of Fructose = 1,017.03 g fructose or 1.017 kg fructose

The answer to the question is d

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"