45) 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 to the water and again brings it to a boil. The
temperature of the salty, boiling water is _ _ _ _ __ _ _ _ _°C.
It is a nice day at sea level so that pressure is 1.00 atm. Assume negligible evaporation of water.
Kb for water is 0.52°C/m.

my book says the answer is: 100.26

how do i solve it?

delta T = i*Kb*m

i = 2
Kb - 0.52
moles NaCl = 58/58.44 = 0.992
m = moles/kg solvent = 0.992/4.01 = 0.247

After finding delta T, add 100 + delta T = boiling temperature.

100.26

100.26

To solve this problem, you need to understand the concept of boiling point elevation. Boiling point elevation is the increase in the boiling point of a solvent when a solute is dissolved in it.

The formula to calculate the boiling point elevation is:

ΔTb = Kb * m

Where:
ΔTb is the change in boiling point (in Celsius).
Kb is the boiling point elevation constant (in Celsius per molal).
m is the molality of the solution (moles of solute per kilogram of solvent).

In this case, George adds 58 g of table salt to 4.01 kg (4001 g) of water. First, convert the mass of the salt and water into moles:

Moles of salt = mass / molar mass
Molar mass of table salt (NaCl) = 58.44 g/mol
Moles of salt = 58 g / 58.44 g/mol = 0.993 mol

Moles of water = mass / molar mass
Molar mass of water = 18.0152 g/mol
Moles of water = 4001 g / 18.0152 g/mol = 222.11 mol

Now, calculate the molality (m) of the solution:

m = moles of solute / mass of solvent (in kg)
m = 0.993 mol / 4.01 kg = 0.247 mol/kg

Finally, substitute the value of molality into the boiling point elevation formula:

ΔTb = Kb * m
ΔTb = 0.52°C/m * 0.247 mol/kg = 0.12844 °C

Add this value to the boiling point of pure water (100 °C) to find the boiling point of the salty water:

Boiling point = 100 °C + 0.12844 °C = 100.13 °C

Therefore, the temperature of the salty, boiling water is 100.13 °C. It is possible that your book may have rounded the answer to 100.26 °C for simplicity or to account for any uncertainties in experimental measurements.