Answer: increase the reactants
Explanation:
Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
Answer:
Procedure (2)
Explanation:
Assume the dialyses come to equilibrium in the allotted times.
Procedure (1)
If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

Procedure (2)
For the first dialysis, the factor is

After a second dialysis, the original concentration of NaCl will be reduced by a factor of

Procedure (2) is more efficient by a factor of
