Question

Enter your answer in the provided box. A solution of 1.55 g of solute dissolved in 25.0 mL of H20 at 25°C has a boiling point of 100.850°C. What is the molar mass of the solute if it is a nonvolatile nonelectrolyte and the solution behaves ideally (d of H20 at 25°C = 0.997 g/mL)? g/mol

Answer 1

Given 1.55 g of solute

25 ml of solvent is water

Temprature T2= 100.850°C

Density of H2O is = 0.997g/ml

Now, from the formula

∆Tb = Kb. m ........ Where m is molality

Kb is constant 0.512

∆Tb is elevation in boiling Point

Now , mass of water = density* volume of water

= 0.997*25 = 24.925 g

Put velues in above formula we get

100.850 - 100 = 0.512 * m

m = 1.660

From above number of moles = mass / molar mass

Hence, molar mass = 1.55/1.660*24.925*10^-3

Molar mass = 1.55/41.3755*10^-3

Molar mass = 37.46 g

Hence molar mass is 37.46 g