3.(12 pts) You have an aqueous solution of sodium acetate (NaCH CO2) that has a

Question

3.(12 pts) You have an aqueous solution of sodium acetate (NaCH CO2) that has a concentration of 3.65 M and a density of 1.16 g/mL. (a) What volume (in mL) of this solution is needed to create 325.0 mL of a 0.025 M sodium acetate solution? (b) what is the mole fraction, of NaCHCO) in this solution? 1. (8 pts) Calculate the freezing point, Trof an aqueous solution prepared by disolving 5,0 grams of 5,.(8 pts) How many grams of aluminum nitrate, AINO) are required to raise the boiling point of 6. (12 pts) Consider each set of substances below. You should assume that all solutions are aqueous and that all solutes are completely dissolved at the indicated concentrations. For each set, circle the substance with the highest value of the indicated parameter (Pap. Tr. T, or osmotic pressure, I1) and underline the solution with the lowest value of that parameter (a) Pap: 0.5 m sucrose 1.5 m sucrose 1.0 m NaCl 1.5 m LisPO 1.0 1(CN)s 2.0 m sucrose 1.0 m MgN2 4.0 m sucrose 1.0NMg(NO3)2 1.0 m sucrose H20 1.5 m KCI 2.0 AClucose (CzH2O1) (b) Tb:1.0 m glucose (c) Tr (d) IT:1.5 MCaClh (CoH1206) 1.5 m CasP2

Explanation / Answer

3. a) Concentration of sodium acetate available = M1 = 3.65 M

Concentration of sodium acetate required = M2 = 0.025 M

Volume required = V2 = 325.0 ml

By using the formula,

M1V1 = M2V2

3.65*V1 = 0.025*325.0

V1 = 2.23 ml

2.23 ml of 3.65 M sodium acetate is required to make 0.025 M sodium acetate.

b) Final concentration of sodium acetate = 0.025 M

volume = 325.0 ml = 0.325 L

Moles of sodium acetate = concentration*volume = 0.025*0.325 = 0.00813 mol

Rest is water. As this quantity is very small, so we can consider that volume of water = 325.0 ml

Density of solution = 1.16 g/ml

Mass of water = volume*density = 325.0*1.16 = 377.0 g

Molar mass of water = 18 g/mol

Moles of water = mass/molar mass = 377.0/18 = 20.94 mol

Mole fraction of sodium acetate = moles of sodium acetate/(moles of sodium acetate+moles of water)

= 0.00813/(20.94+0.00813) = 3.88*10-4

4. Mass of lithium carbonate = 5.0 g

Molar mass of Li2CO3 = 74 g/mol

Moles of Li2CO3 = 5.0/74 = 0.068 mol

Mass of water = 40.0 g = 0.040 kg

Molality of solution = moles of Li2CO3/mass of water = 0.068/0.040 = 1.7 m

Freezing point of solution = Tf = ?

Freezing point of water = Tfo = 0 oC

Cryoscopic constant for water = Kf = 1.85

Tfo-Tf = iKfm

0-Tf = 3*1.85*1.7

Tf = -9.4 oC

Freezing point of solution = -9.4 oC

5) Elevation in boiling point of water = 105.0 - 100.0 = 5.0 oC

vant hoff factor for Al(NO3)3 = 4

Kb for water = 0.512

5.0 = 4*0.512*m

m = 2.44

Molality of solution = 2.44

Mass of water = 25.0 g = 0.025 kg

Moles of Al(NO3)3 = molality*mass of water = 2.44*0.025 = 0.061 mol

Molar mass of Al(NO3)3 = 213 g/mol

Mass of Al(NO3)3 in solution = mol*molar mass = 0.061*213 = 13.00 g

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