0 votes
57 views
in Physical Chemistry by (450 points)
A chemistry student is trying to determine the boiling point of a solution containing 30 grams of glucose (C6H12O6) dissolved in 100 grams of water. Given that the boiling point of pure water is 100°C and the elevation constant for water is 0.512°C/m, what is the boiling point of the solution?

1 Answer

0 votes
by (390 points)
To determine the boiling point of the solution, we can use the formula for boiling point elevation:Tb = Kb * molality * iwhere Tb is the change in boiling point, Kb is the boiling point elevation constant for water  0.512C/m , molality is the molality of the solution, and i is the van't Hoff factor  number of particles the solute dissociates into, which is 1 for glucose as it does not dissociate in water .First, we need to find the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. 1. Calculate the moles of glucose:Molar mass of glucose  C6H12O6  =  6 * 12.01  +  12 * 1.01  +  6 * 16.00  = 72.06 + 12.12 + 96.00 = 180.18 g/molMoles of glucose =  30 g  /  180.18 g/mol  = 0.1666 mol2. Convert the mass of water to kilograms:100 g = 0.1 kg3. Calculate the molality of the solution:Molality =  0.1666 mol  /  0.1 kg  = 1.666 mol/kg4. Calculate the boiling point elevation:Tb = Kb * molality * iTb =  0.512C/m  *  1.666 mol/kg  * 1 = 0.853C5. Determine the boiling point of the solution:Boiling point of solution = boiling point of pure water + TbBoiling point of solution = 100C + 0.853C = 100.853CThe boiling point of the solution is approximately 100.853C.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...