If sucrose (C12H22O11) are dissolved in water to make a 0.821m solution, what is the boiling point of the resulting solution? Kb for water = 0.512°C/m. ΔTb = Kb x m

Question

a
99.626°C
b
101.64°C
c
100.73°C
d
100.42°C

Answer 1

To find the boiling point of the solution, we can use the formula:

\[ \Delta T_b = K_b \times m \]

Where:

• \(\Delta T_b\) is the boiling point elevation
• \(K_b\) is the ebullioscopic constant of the solvent (water in this case, \(K_b = 0.512°C/m\))
• \(m\) is the molality of the solution (0.821 m)

First, we can calculate \(\Delta T_b\):

\[ \Delta T_b = 0.512 , °C/m \times 0.821 , m = 0.420512 , °C \]

Now, we need to calculate the new boiling point of the solution. The normal boiling point of water is 100°C, so we add the change in boiling point to this:

\[ \text{New boiling point} = \text{Normal boiling point} + \Delta T_b \] \[ \text{New boiling point} = 100°C + 0.420512°C \approx 100.42°C \]

Now, rounding to two decimal places:

The new boiling point of the solution is approximately 100.42°C, which corresponds to option d.