Answer
a. Freezing point: -1.08 $^oC$
b. Freezing point: -2.25 $^o C$
c. Freezing point: -11 $^oC$
Work Step by Step
a.
1. Calculate the $\Delta T_f$
1 mole of particles = 1.86 $^o C
$
$lactose$ is a nonelectrolyte: 1 mole $lactose$ = 1 mole particles
$$0.580 \space moles \space lactose \times \frac{1 \space mole \space particles}{1 \space mole \space lactose} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 1.08 \space ^oC$$
2. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 1.08 \space ^o C = -1.08 \space ^o C$$
b.
1. Find the molar mass for this compound:
$ KCl $ : ( 35.45 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 74.55 g/mol
2. Calculate the $\Delta T_f$
1 mole of particles = 1.86 $^o C
$
$KCl$ is an electrolyte that produces 2 ions: 1 mole $KCl$ = 2 moles particles
$$45.0 \space g \space KCl \times \frac{1 \space mole KCl}{74.55 \space g \space KCl} \times \frac{2 \space mole s \space particles}{1 \space mole \space KCl} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 2.25 \space ^oC$$
3. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 2.25 \space ^o C = -2.25 \space ^o C$$
c.
1. Calculate the $\Delta T_f$
1 mole of particles = 1.86 $^o C
$
$K_3PO_4$ is an electrolyte that produces 4 ions: 1 mole $K_3PO_4$ = 4 moles particles
$$1.5 \space moles \space K_3PO_4 \times \frac{4 \space mole s \space particles}{1 \space mole \space K_3PO_4} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 11 \space ^oC$$
2. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 11 \space ^o C = -11 \space ^o C$$
