Question

1.What is the boiling point of a solution of 3.00g of ascorbic acid(molar mass=176.12) in 50.0g of acetic acid(KB=3.07 degrees celcius?

2.How many moles of particles are present in 1.0L of each of the following solutions:

A. .150M NaCH3CO2

B. .02M (NH4)3PO3

Answer 1

1.

Molecular weight (or mass) of ascorbic acid is 176.12 g/mole

(3g)/(176.12 g/mole) = 0.017 mole

concentration = (0.017 mole)/(50 g) = (0.017 mole)/[(50 g)/(1000 g/kg)] = 0.341 moles/kg

Concentration in terms of molality = 0.341 = c_m

?T = (Kb)(c_m) = (3.07)(0.341 moles/kg) = 1.047 degree celcius

change in boiling point is 1.047 degree celcius.

boiling point acetic acid should be given.

2.

molarity = moles/volume

since volume is 1L so molarity = moles

A. moles - .150 moles of compound

moles of Na = 0.150

moles of C = 0.150*2 = 0.300

moles of H = 0.150*3 = 0.450

moles if O = 2*.150 = 0.300