Pure water boils at 100ºC (212ºF). When a substance is added to the water the boiling point increases due to what is referred to as the boiling point elevation. (a) What is the molecular mass of sucrose (C₁₂H₂₂O₁₁)? (b) If you add 100 g of sucrose to water, how many moles of sucrose would you then have? (c) If you add the sucrose to 500 g of water, by what fraction would you divide? (Hint: boiling point elevation for water is 0.521ºC for each mole of solute dissolved in 1000 g of solvent) (d) What will the boiling point elevation be if 100 g of sucrose is added to 500 g of water? (boiling point elevation for water is 0.521ºC for each mole of solute dissolved in 1000 g of solvent) (e) What will the boiling point be if 100 g of sucrose is added to 500 g of water?

mols sucrose = grams sucrose/molar mass sucrose.

molality sucrose = mols sucrose/kg solvent.

delta T = Kb*molality

boiling point of soln = 100 + delta T.

To answer these questions, we'll need to use the concepts of molecular mass, moles, and boiling point elevation. Let's go step by step:

(a) To find the molecular mass of sucrose (C₁₂H₂₂O₁₁), we need to calculate the sum of the atomic masses of each element in the chemical formula. The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are approximately 12, 1, and 16 grams per mole, respectively.

Molecular mass of sucrose = (12 * 12) + (1 * 22) + (16 * 11)
= 144 + 22 + 176
= 342 grams per mole

So, the molecular mass of sucrose is 342 grams per mole.

(b) To calculate the number of moles of sucrose when 100 g is added to water, we need to divide the mass of sucrose by its molecular mass:

Number of moles of sucrose = Mass of sucrose / Molecular mass of sucrose
= 100 g / 342 g/mol
≈ 0.292 moles

So, 100 g of sucrose is approximately equal to 0.292 moles.

(c) The fraction by which you would divide when you add the sucrose to 500 g of water can be calculated using the mass of the solvent (water) and density formula. The density of water is approximately 1 g/mL, so 500 g of water is equal to 500 mL.

Fraction = Mass of solvent / (Mass of solvent + Mass of solute)
= 500 g / (500 g + 100 g)
= 500 g / 600 g
≈ 0.833

So, when you add sucrose to 500 g of water, you would divide it by approximately 0.833.

(d) The boiling point elevation for water is 0.521ºC for each mole of solute dissolved in 1000 g of solvent. To find the boiling point elevation when 100 g of sucrose is added to 500 g of water, we first need to calculate the number of moles of sucrose using the same formula as in (b):

Number of moles of sucrose = Mass of sucrose / Molecular mass of sucrose
= 100 g / 342 g/mol
≈ 0.292 moles

The boiling point elevation can be calculated as follows:

Boiling point elevation = Number of moles of solute * Boiling point elevation constant
= 0.292 moles * 0.521ºC/mole
≈ 0.152ºC

So, when 100 g of sucrose is added to 500 g of water, the boiling point elevation would be approximately 0.152ºC.

(e) To find the new boiling point when 100 g of sucrose is added to 500 g of water, we need to add the boiling point elevation to the boiling point of pure water.

Boiling point = Boiling point of pure water + Boiling point elevation
= 100ºC + 0.152ºC
≈ 100.152ºC

So, the boiling point would be approximately 100.152ºC when 100 g of sucrose is added to 500 g of water.

Remember, these calculations are approximate values based on the given information and assumptions.