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

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

The correct answer is b.

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 =

number of moles = 0.17 moles

molality =

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.

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