a) The freezing point will decrease the most (compared to parts "b" and "c". If the molality of the solute is m, the lowering of the freezing point will be
deltaT = 2•Kf•m (the effective molality is 2m)
b) The lowering of the freezing point will be
deltaT = Kf•m (compare this to part "a").
c) If the molality of C is m,
deltaT = Kf•m (compare this to part "a" and "b")
What effect do each of these have on the freezing point of the solvent?
a. a nonvolatile solute that dissociates in the solvent
b. a volatile solute that does not dissociate
c. two solutes that react according to the equation:
A+B---> C
Please explain why.
2 answers
Comment of Part C:
Normally the effective molality when we have two solutes would be the SUM of the two molalities:
m(effective) = m(A) + m(B)
However since A and B combine to form C, if we use equal moles of the two solutes, the effective molality will be equal to the molality of ONE of them which is equal to the final molality of C. If we mix A and B at a mole ratio not equal to 1, things get more complicated since the effective molality equals the molality of C plus the final molality of the excess reagent.
That is not a well thought out question unless it was constructed to stimulated thinking about all possible scenarios.
Normally the effective molality when we have two solutes would be the SUM of the two molalities:
m(effective) = m(A) + m(B)
However since A and B combine to form C, if we use equal moles of the two solutes, the effective molality will be equal to the molality of ONE of them which is equal to the final molality of C. If we mix A and B at a mole ratio not equal to 1, things get more complicated since the effective molality equals the molality of C plus the final molality of the excess reagent.
That is not a well thought out question unless it was constructed to stimulated thinking about all possible scenarios.
2 answers