18g of glucose is dissolved in 1kg of water in a saucepan.at what temperature will the water boil at 1.013 bar

To determine the boiling point of water when 18 grams of glucose is dissolved in 1 kilogram of water, we need to consider the concept of boiling point elevation. This phenomenon occurs when a solute is added to a solvent, resulting in an increase in the boiling point of the solvent. Let's break this down step by step.

Understanding Boiling Point Elevation

The boiling point elevation can be calculated using the formula:

ΔT_b = i * K_b * m

• ΔT_b = change in boiling point
• i = van 't Hoff factor (number of particles the solute breaks into)
• K_b = ebullioscopic constant of the solvent (for water, K_b is approximately 0.512 °C kg/mol)
• m = molality of the solution (moles of solute per kg of solvent)

Calculating Molality

First, we need to find the molality of the glucose solution. Glucose (C₆H₁₂O₆) has a molar mass of about 180 g/mol. To find the number of moles of glucose in 18 grams, we use the formula:

moles of glucose = mass (g) / molar mass (g/mol)

Substituting the values:

moles of glucose = 18 g / 180 g/mol = 0.1 moles

Finding Molality

Since we have 1 kg of water, the molality (m) is simply:

m = moles of solute / kg of solvent = 0.1 moles / 1 kg = 0.1 mol/kg

Determining the Van 't Hoff Factor

For glucose, which does not dissociate in solution, the van 't Hoff factor (i) is 1. This means:

i = 1

Calculating Boiling Point Elevation

Now we can plug these values into the boiling point elevation formula:

ΔT_b = i * K_b * m

Substituting the known values:

ΔT_b = 1 * 0.512 °C kg/mol * 0.1 mol/kg = 0.0512 °C

Finding the New Boiling Point

The normal boiling point of water at 1.013 bar (which is approximately 1 atm) is 100 °C. To find the new boiling point, we add the boiling point elevation to the normal boiling point:

New boiling point = 100 °C + 0.0512 °C = 100.0512 °C

Final Result

Therefore, when 18 grams of glucose is dissolved in 1 kilogram of water, the boiling point of the solution at 1.013 bar will be approximately 100.05 °C.