Answer
a. Freezing point = - 2.53 $^o C$
b. Freezing point = - 15.6 $^o C$
c. Freezing point = - 5.54 $^o C$
Work Step by Step
a.
1. Calculate the $\Delta T_f$
1 mole of particles = 1.86 $^o C
$
$CH_3OH$ is a nonelectrolyte: 1 mole $CH_3OH$ = 1 mole particles
$$1.36 \space moles \space CH_3OH \times \frac{1 \space mole \space particles}{1 \space mole \space CH_3OH} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 2.53 \space ^oC$$
2. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 2.53 \space ^o C = -2.53 \space ^o C$$
b.
1. Find the molar mass for this compound:
$ C_3H_8O_2 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 3 )+ ( 16.00 $\times$ 2 )= 76.09 g/mol
2. Calculate the $\Delta T_f$
1 mole of particles = 1.86 $^o C
$
$C_3H_8O_2$ is a nonelectrolyte: 1 mole $C_3H_8O_2$ = 1 mole particles
$$640. \space g \space C_3H_8O_2 \times \frac{1 \space mole C_3H_8O_2}{76.09 \space g \space C_3H_8O_2} \times \frac{1 \space mole \space particles}{1 \space mole \space C_3H_8O_2} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 15.6 \space ^oC$$
3. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 15.6 \space ^o C = -15.6 \space ^o C$$
c.
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
$$111 \space g \space KCl \times \frac{1 \space mole KCl}{74.55 \space g \space KCl} \times \frac{2 \space mole s \space particles}{1 \space mole \space KCl} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 5.54 \space ^oC$$
3. The freezing point of water is $0$ $^o C$, calculate the freezing point of the solution.
$$T_{solution} = T_{water} - \Delta T_f$$
$$T_{solution} = 0 \space ^oC - 5.54 \space ^o C = -5.54 \space ^o C$$
