1. Calculate the boiling point elevation of .100kg of water containing .010 mol
ID: 798791 • Letter: 1
Question
1. Calculate the boiling point elevation of .100kg of water containing .010 mol of NaCl, 0.02 mol of Na2SO4 and .03 mol of MgCL2 assuming complete dissociation of electrolytes (Kb = .51degreesC/m)
2. calculate the vapor pressure of water above a solution prepare by dissolving 22.5 g of lactose (C12H22O11) in 200 g of water if the vapor pressure of pure water is 187.5 torr.
3, A 100ml sample of sodium bromide contained 104.62 g of sodium bromide. Upon evaporation, 6.280 g of residue was obtained. From these data, calculate each of the following for the solution: a) density of the solution b) percent by mass of sodium bromide in the solution c) concentration of sodium bromide in grams per 100 ml of water. d) grams of sodium bromide per liter of solution e) molarity of the solution
4. An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient water to form 250 ml of solution. The solution has an osmotic pressure of 1.2 atm at 25degreesC. WHat is the molar mass (g/mole) of the compound?
5. The peroxydisulfate ion ( S2O82-) reacts with the iodide ion in aqueous solution via the reaction:
S2O8- (aq) + 3I this is an arrow 2SO4 (aq) + I3- (aq)
An aquaeous solution containing 0.050 M of S2O82- ion and 0.072 M of I- is prepared. and the progress of the reaction followed by measuring [I-]. the data obtained is given in the table below
Time (s) 0.000 400.0 800.0 1200.0 1600.0
[I-] (M) 0.072 0.057 0.057 0.037 0.029
A) the average rate of disappearance of I- between 400.0 s and 800.0 s is M/s.
B) the average rate of disapearance of I- in the initial 400.0s is M/s.
6. The value of Keq for the equilibrium
H2 (g) + I2 ( g) arrow from lower to higher 2 HI (g)
is 794 at 25 celcuis degrees. At this temperature, what is the value of Keq for the equilibrium below?
HI (g) square 1/2 H2 (g) + 1/2 I2 (g)
Explanation / Answer
delta Tb = Kb * b solute * i
i = 2 for NaCl , 3 for Na2SO4 ,3 for MgCl2 since complete dissociation
Kb = 0.51degC/m
delta Tb= Kb * { [(0.010 * 2) + (0.02 * 3 ) + (0.03 * 3)] / [3]}
=0.51 *{ 0.02 + 0.06 + 0.09 / 3 }
=0.51 * 0.17/3
= 0.17 * 0.17
=0.0289
elevation in bolioling point = 0.0289
2. Molecular weight of lactose :-342.3 g/mol
moles of lactose M1 = wt /mol.wt = 0.06573
Molecular weight of water = 18
moles of water M2 = 200 / 18 = 11.11
Therefore mole fraction of lactose = M1 /(M1 + M2) = 5.882 * 10^-3
Molefraction of water = 1 - lactose mole fraction = 0.9941
Therefore vapor pressure of water = molefracction of water * partial presssure of water
vapor pressure of pure water is 187.5 torr
vapor pressure = 187.5 * 0.9941 = 186.397
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