Answer
a. 100.495 $^o C$
b. 102.8 $^o C$
c. 100.74 $^o C$
Work Step by Step
a.
1. Find the molar mass for this compound:
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
2. Calculate the $\Delta T_b$
1 mole of particles = 0.51 $^o C
$
$C_6H_{12}O_6$ is a nonelectrolyte: 1 mole $C_6H_{12}O_6$ = 1 mole particles
$$175 \space g \space C_6H_{12}O_6 \times \frac{1 \space mole C_6H_{12}O_6}{180.16 \space g \space C_6H_{12}O_6} \times \frac{1 \space mole \space particles}{1 \space mole \space C_6H_{12}O_6} \times \frac{0.51 \space ^o C}{1 \space mole \space particles} = 0.495 \space ^oC$$
3. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_f$$
$$T_{solution} = 100 \space ^oC + 0.495 \space ^o C = 100.495\space ^o C$$
b.
1. Calculate the $\Delta T_b$
1 mole of particles = 0.51 $^o C
$
$CaCl_2$ is an electrolyte that produces 3 ions: 1 mole $CaCl_2$ = 3 moles particles
$$1.8 \space moles \space CaCl_2 \times \frac{3 \space mole s \space particles}{1 \space mole \space CaCl_2} \times \frac{0.51 \space ^o C}{1 \space mole \space particles} = 2.8 \space ^oC$$
2. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_b$$
$$T_{solution} = 100 \space ^oC + 2.8 \space ^o C = 102.8 \space ^o C$$
c.
1. Find the molar mass for this compound:
$ LiNO_3 $ : ( 6.941 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 68.95 g/mol
2. Calculate the $\Delta T_b$
1 mole of particles = 0.51 $^o C
$
$LiNO_3$ is an electrolyte that produces 2 ions: 1 mole $LiNO_3$ = 2 moles particles
$$50.0 \space g \space LiNO_3 \times \frac{1 \space mole LiNO_3}{68.95 \space g \space LiNO_3} \times \frac{2 \space mole s \space particles}{1 \space mole \space LiNO_3} \times \frac{0.51 \space ^o C}{1 \space mole \space particles} = 0.740 \space ^oC$$
3. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_f$$
$$T_{solution} = 100 \space ^oC + 0.740 \space ^o C = 100.74 \space ^o C$$
