1) Calculate the molar mass of a compound if a solution of 12.0g dissolved in 80

Question

1) Calculate the molar mass of a compound if a solution of 12.0g dissolved in 80.0g of water freezes at -1.94 C. (Kf water= 1.86 C/m) Answer is supposed to be 144g/mol but how do I get there?

2) Calculate the total vapor pressure of a solution made by dissolving 25.0g of glucose (C6H12O6) in 215g of water at 50 C.

(MM C6H12O6 = 180.2 g/mol, MM H2O = 18.02 g/mol. Vapor pressure of H2O @ 50 C= 92.5 torr) ANS: 91.4 torr

3) Calculate the boiling point of a solution made by dissolving 1.00g of glycerin, C3H8O3, in 54.0 g of water.

(Kb for H2O = 0.512 C/m)

4)Lysozome is an enzyme used to cleave cell walls. A solution made by dissolving 0.0750g of Lysozome in 100.0 mL results in an osmotic pressure of 1.32x10^-3 atm at 25 C. Calculate the molar mass of lysozome.

Any help will be appreciated! thanks! show the steps please, I'm really trying to understand these.

Explanation / Answer

delta t = m x Kf x i

if we assume only 1 particle in solution we have

m = 1.94 / 1.86 = 1.043 moles of the unknown per kg of water

we also have 12.0 g of compound in 80.0 g of water or 12.0 x 1000 / 80 g / kg of water

= 150 g per kg of water

so 1.043 moles = 150 g of compound in 1 kg of water

mass of 1 mole = 150 g / 1.043 moles = 144 g / mole // molar mass

2)vapor pressure = xf * P

vapour pressure =[ 11.94/12.077]* 92,5 = 91.4 torr

3)] find elevation in bolilng poitn

tb = (.512 * 1)/92*.054

= 0.01 *C

=1.32x10^-3= .0821 * .075*298/0.1 *m

m = 13901gms

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