Let's solve this two-part problem step by step.
(a) Difference in Boiling Points:
To determine the difference in boiling points between a 7.0 m (molal) glucose solution and pure water, we calculate the boiling point elevation. The boiling point elevation can be found using the formula:
[tex]\[ \Delta T_b = i \times K_b \times m \][/tex]
Where:
- [tex]\( \Delta T_b \)[/tex] is the boiling point elevation.
- [tex]\( i \)[/tex] is the van't Hoff factor. For glucose, which is a non-electrolyte, [tex]\( i = 1 \)[/tex].
- [tex]\( K_b \)[/tex] is the ebullioscopic constant of the solvent (water in this case), given as [tex]\( 0.512 \, ^{\circ}C \cdot kg/mol \)[/tex].
- [tex]\( m \)[/tex] is the molality of the solution, given as [tex]\( 7.0 \, mol/kg \)[/tex].
Using the provided information, we find:
[tex]\[ \Delta T_b = 1 \times 0.512 \, ^{\circ}C \cdot kg/mol \times 7.0 \, mol/kg = 3.584 \, ^{\circ}C \][/tex]
Thus, the difference in boiling points is [tex]\( \Delta T_b \)[/tex]:
[tex]\( 3.584 \, ^{\circ}C \)[/tex]
(b) Boiling Point of the 7.0 m Glucose Solution:
Next, we determine the boiling point of the glucose solution. To do this, we add the boiling point elevation ([tex]\( \Delta T_b \)[/tex]) to the boiling point of pure water.
The boiling point of pure water is [tex]\( 100 \, ^{\circ}C \)[/tex]. Adding the boiling point elevation to this value gives:
[tex]\[ T_{solution} = 100 \, ^{\circ}C + 3.584 \, ^{\circ}C = 103.584 \, ^{\circ}C \][/tex]
Therefore, the boiling point of the 7.0 m glucose solution is:
[tex]\( 103.584 \, ^{\circ}C \)[/tex]
In summary:
(a) The difference in boiling points: [tex]\( 3.584 \, ^{\circ}C \)[/tex]
(b) The boiling point of the 7.0 m glucose solution: [tex]\( 103.584 \, ^{\circ}C \)[/tex]
