How many grammes of Sucrose will be added to 0.600kg of water to elevate the Boiling point by 5.0 degrees Celsius

To determine the amount of sucrose needed to elevate the boiling point of water by a certain number of degrees Celsius, we can use the formula:

ΔTb = Kbm

Where:
ΔTb is the boiling point elevation
Kb is the molal boiling point elevation constant for water (0.512 °C/m)
m is the molality (moles of solute per kilograms of solvent)

First, we need to calculate the molality (m) of the solution:

Given:
Mass of water (solvent) = 0.600 kg
Change in boiling point (ΔTb) = 5.0 °C
Kb (molal boiling point elevation constant for water) = 0.512 °C/m

Step 1: Convert the mass of water from kilograms to grams.
0.600 kg = 600 g

Step 2: Calculate the moles of sucrose needed to elevate the boiling point.
m = (ΔTb) / (Kb)
m = 5.0 °C / 0.512 °C/m
m ≈ 9.77 mol/kg

Step 3: Convert the molality (m) into moles of sucrose.
m = (moles of solute) / (mass of solvent in kg)
9.77 mol/kg = (moles of sucrose) / 0.600 kg

Rearranging the equation, we can determine the moles of sucrose:
(moles of sucrose) = m * (mass of solvent in kg)
(moles of sucrose) ≈ 9.77 mol/kg * 0.600 kg
(moles of sucrose) ≈ 5.86 mol

Step 4: Convert moles of sucrose into grams.
1 mole of sucrose = 342.3 g (molar mass of sucrose)
5.86 mol * 342.3 g/mol ≈ 2005 g

Therefore, approximately 2005 grams of sucrose should be added to 0.600 kg of water to elevate the boiling point by 5.0 degrees Celsius.