Question

I will rate, please answer.

The density of an aqueous solution of C2H602(l) is 1.48 g/ml. The molarity of the solution is 2.84. Find the freezing point of the solution and the vapor pressure of the solution at 25 C. (Kf of H20 = 1.86 c/m), VP of H20 at 25 C = 18.4 torr)

Answer 1

Density of the solution is 1.48g/ml

which means 1mL of the solution weight 1.48 g or 1 L of the solution weights 1480g

The molarity of the solution is 2.84 which means it is 2.84 moles/L

2.84 moles x 62 g/mol = 176g

Since 1 L has a mass of 1480g the difference is the solvent weight

1480 - 176 = 1304 g

Something is wrong here as per the data for the density, molecular weight and molarity provided a 1 L solution of a substance in water if it has such a high density cannot be so low molarity with a molecular weight which is so small.

Still I will tell you how to proceed with the calculation

freezing point depresssion is

ΔTf=−Kf×m

here m is molality not molarity

molality is defines as moles /Kg of solvent

molality is moles /L

if we assume both are same

ΔTf=−Kf×m

ΔTf=−1.86 x 2.84

ΔTf=−5.28

so the freezing point decreases by 5.28 degrees.

Vapour pressure of a solution is given by the formula

xsolv = x_solv is the mole fraction of the solvent.the fraction of the total number of moles present which is solvent.

so here we need to know the moles of the water versus the moles of the material. Where the total amount of water in a solution is coming out wrong