Answer
a. 101.08 $^o C$
b. 100.16 $^o C$
c. 101.75 $^o C$
Work Step by Step
a.
1. 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
$$2.12 \space moles \space C_6H_{12}O_6 \times \frac{1 \space mole \space particles}{1 \space mole \space C_6H_{12}O_6} \times \frac{1.86 \space ^o C}{1 \space mole \space particles} = 1.08 \space ^oC$$
2. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_b$$
$$T_{solution} = 100 \space ^oC + 1.08 \space ^o C = 101.08 \space ^o C$$
b.
1. Find the molar mass for this compound:
$ C_{12}H_{22}O_{11} $ : ( 1.008 $\times$ 22 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 11 )= 342.30 g/mol
2. Calculate the $\Delta T_b$
1 mole of particles = 0.51 $^o C
$
$C_{12}H_{22}O_{11}$ is a nonelectrolyte: 1 mole $C_{12}H_{22}O_{11}$ = 1 mole particles
$$110. \space g \space C_{12}H_{22}O_{11} \times \frac{1 \space mole C_{12}H_{22}O_{11}}{342.30 \space g \space C_{12}H_{22}O_{11}} \times \frac{1 \space mole \space particles}{1 \space mole \space C_{12}H_{22}O_{11}} \times \frac{0.51 \space ^o C}{1 \space mole \space particles} = 0.164 \space ^oC$$
3. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_f$$
$$T_{solution} = 100 \space ^oC + 0.164 \space ^o C = 100.16 \space ^o C$$
c.
1. Find the molar mass for this compound:
$ NaNO_3 $ : ( 22.99 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 85.00 g/mol
2. Calculate the $\Delta T_b$
1 mole of particles = 0.51 $^o C
$
$NaNO_3$ is an electrolyte that produces 2 ions: 1 mole $NaNO_3$ = 2 moles particles
$$146 \space g \space NaNO_3 \times \frac{1 \space mole NaNO_3}{85.00 \space g \space NaNO_3} \times \frac{2 \space mole s \space particles}{1 \space mole \space NaNO_3} \times \frac{0.51 \space ^o C}{1 \space mole \space particles} = 1.75 \space ^oC$$
3. The boiling point of water is $100$ $^o C$, calculate the boiling point of the solution.
$$T_{solution} = T_{water} + \Delta T_f$$
$$T_{solution} = 100 \space ^oC + 1.75 \space ^o C = 101.75 \space ^o C$$
