Question

You add 31.65 g of a non-electrolyte solute to 220.0 mL of water to cook Mac and Cheese. How much hotter will the water be? The molar mass of the solute is 180.18 g/mol and the density of water is 1.0 g/m

a) 0.004072^{\circ}C
b) 100.4^{\circ}C
c) 73.37^{\circ}C
d) 0.4072^{\circ}C

Answer 1

100.4 degree celsius is the temperature of the water having non-electrolyte.

The correct answer is b.

Step-by-step explanation:

Data given:

mass of non-electrolyte solute = 31.65 grams

volume of water = 220 ml or 0.22 litres

molar mass of solute = 180.18 grams/mole

density of water = 1 gm/mole

Boiling point of water = 100 degrees celsius

molality = ?

Kb for water = 0.51

boiling point elevation or temperature of hot water,T =?

Formula used:

T = mKb equation 1

number of moles of non electrolyte =
`(31.65)/(180.18)`

number of moles = 0.17 moles

molality =
`(0.17)/(0.22)`

molality = 0.77 M

putting the values in equation 1

T = 0.77 X 0.51

T = 0.392 degree celsius. is the elevation in temperature when solute is added.

Temperature of the hot water = 100 +0.392

= 100.392 degree celsius.