Date_ Experiment 1: The Preparation and Dilution of a 0.10 Molar Solution I. Pur

Question

Date_ Experiment 1: The Preparation and Dilution of a 0.10 Molar Solution I. Purpose: Il. Test Data Part 1: Preparation of a 0.10 M Sucrose Solution 1. Mass of empty 100 mL graduated cylinder Uaram 2. Mass of Kook-Aide powder (sucrose) 3. volume of solution in 100 mL graduated cylinder LOON 4. Mass of solution in 100 mL graduated cylinder .arams Part 2: Serial Dilutions of a Sucrose Solution 3 Harams I. Test Tube #2 50mL Volume of solution from test tube #1 Volume of distilled water 2. Test Tube #3 Volume of solution from test tube #2 50 mL Volume of distilled water

Explanation / Answer

III) calculations:

1) mass of solute=3.4g

moles of solute (sucrose)=mass of solute/molar mass of sucrose=3.4g/342.3g/mol=0.00993 mol

2) Solution

1) volume of sucrose solution(in L)=100 ml

2) Mass of sucrose solution=184.9 g

3)Concentration of Sucrose

a) Mass/volume % of solution=(mass of sucrose/volume of solution)*100=(3.4g/100ml)*100=3.4%

b)Mass/mass% of solution=(mass of sucrose/mass of solution)*100=(3.4/184.9)*100=1.8%

c) Molarity of the solution=mol of sucrose/volume of solution=0.00993 mol/100ml=0.00993 mol/0.1L=0.0993 M

4) a) Test tube #1

Molarity of the solution=0.0993 M

b) test tube#2

volume of solution from test tube 1=5.0ml

volume of distilled water=5.0ml

total volume of solution 2=5.0ml+5.0ml=10.0ml

Molarity of solution 2=Molarity of 1*(volume of sol1/vol of sol2)=0.0993 M*(5.0ml/10.0ml)=0.0496M

c) Test tube #3

volume of solution from test tube 2=5.0ml

volume of distilled water=5.0ml

total volume of solution 3=5.0ml+5.0ml=10.0ml

Molarity of solution 3=Molarity of 2*(volume of sol1/vol of sol2)=0.0496M*(5.0ml/10.0ml)=0.0248 M

d)

Test tube #4

volume of solution from test tube 3=5.0ml

volume of distilled water=5.0ml

total volume of solution 3=5.0ml+5.0ml=10.0ml

Molarity of solution 4=Molarity of 3*(volume of sol1/vol of sol2)=0.0248M*(5.0ml/10.0ml)=0.0124 M

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