The freezing point of 42.19 g of a pure solvent is measured to be 43.17 ºC. When 2.03 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 40.32 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.67 ºC·kg solvent/mol solute).What is the molality of the solution? m
How many moles of solute are present? mol
What is the molecular weight of the solute? g/molIf the freezing point depression constant of the pure solvent is 7.66 ºC·kg solvent/mol solute calculate the molar mass of the solute. g/mol

The molality of the solution is 0.3716 mol/kg

The number of moles of solute is 0.0157 mol

The molecular weight of the solute is 129.30 g/mol

The molar mass of the solute is 129.32 g/mol

Explanation:

m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg

Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol

Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol

When Kf = 7.66 °C.kg/mol

Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.