Question

3) The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of atm. A) 0.790 B) 0.498 C) 0.367 D) 0.413 E) 0.0918 4) George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt (NaCl) to the water and again brings it to a boil. The temperature of the salty, boiling water is
please show all work

Answer 1

Q3. Given : moles of ethanol = 90.0 mmol

moles of naphthalene = 10.0 mmol

Total moles = (moles of ethanol) * (moles of naphthalene)

Total moles = (90.0 mmol) + (10.0 mmol)

Total moles = 100.0 mmol

mole fraction ethanol = (moles of ethanol) / (Total moles)

mole fraction ethanol = (90.0 mmol) / (100.0 mmol)

mole fraction ethanol = 0.900

vapor pressure of solution = (mole fraction ethanol) * (vapor pressure of pure ethanol)

vapor pressure of solution = (0.900) * (0.459 atm)

vapor pressure of solution = 0.413 atm

Q4.

Given : mass NaCl = 58 g

moles NaCl = (mass NaCl) / (molar mass NaCl)

moles NaCl = (58 g) / (58.44 g/mol)

moles NaCl = 0.9925 mol

molality NaCl = (moles NaCl) / (mass of water in kg)

molality NaCl = (0.9925 mol) / (4.01 kg)

molality NaCl = 0.2475 m

increase in boiling point = (i) * (K_b) * (m)

where i = van't Hoff factor = 2 (for NaCl)

K_b = 0.512 ^oC/m (for water)

m = molality NaCl = 0.2475 m

Substituting the values,

increase in boiling point = (2) * (0.512 ^oC/m) * (0.2475 m)

increase in boiling point = 0.25 ^oC

boiling point of solution = (boiling point of pure water) + (increase in boiling point)

boiling point of solution = (100.00 ^oC) + (0.25 ^oC)

boiling point of solution = 100.25 ^oC

Answer 2

1. correct option is c

2. bp is 100.25C