Question

Number 2

Chemistry 1B-082 ,20IS 10:00 AM NO books, notes, other references Chapter 11 1(6) Arrange the following in order of increasing solubility in water AND identify the strongest would exist between water and the solute in each aqu e strongest type would exist between water and the solute in each aqueous solution. CH3OH, SF2, HCL, c C02 HC <C do dipole 2(5) a) How would you prepare 500.00 grams of a 30.0% aqueous sodium, carbonate solution? b) What would be the molality of the solution? 300% Na,D, 105.19 Na CO c) What would be the freezing point of the solution? (Kbp for water is 0.512 °C/m) 4(3) What would be the osmotic pressure of 500. mLofa 0.100MK3PO4 solution? \ -) T-(4)( 0-100 y()( 0.0%20e ) ( 273 Y)

Answer 1

Number 2:
Part a:
Total mass = 500 gm
Mass % of sodium carbonate = 30 %
Use:
Mass percent = mass of sodium carbonate *100 / Total mass
30 = mass of sodium carbonate *100 / 500
mass of sodium carbonate = 150 gm
Mass of water = 500-150 = 350 gm
Since density of water is 1gm/ml, Above solution can be prepared by dissolving 150 gm of sodium carbonate in 350 ml of water.

Part B:
Molar mass of sodium carbonate = 105.99 gm = 106 gm
mass of sodium carbonate = 150 gm
Mass of water/solvent = 350 gm = 0.35 Kg
Number of moles of sodium carbonate = mass of sodium carbonate /Molar mass of sodium carbonate
           = 150/106
           =1.415 mol
Molality = Number of moles of solute / mass of solvent in Kg
                = 1.415 / 0.35
                  = 4.043 m

Part C:
depression in freezing point = Kbp * molality
    =0.512*4.043
   = 2.07 oC
New Freezing point = 0 - 2.07 oC
   = -2.07 oC