Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.
So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
Use Raoult's Law:
Psolution = (χsolvent) (P°solvent)
24.90 = (x) (25.756)
x = 0.966765 (this is the solvent mole fraction)
χsolute = 1 - 0.966765 = 0.033235
χsolute = 0.03324 (to four sig figs)
Answer:
lesser the molar mass of the gas higher the no. of moles included in a certain mass sample. ie at STP more volume is required for the gas having less molar mass.
He has the smallest molar mass.
Therefore bag of He is the biggest.
