Question

Water Conductance Lab (Redone)

**Known Brine Solution is (0.100 M) NaCl

Part 1: Conductance of Known Solutions

Brine Solutions	Conductivity	Temperature (with thermometer)
#1 (100 ppm)	0.01 µS	25°C
#2 (500 ppm)	0.07 µS	25°C
#3 (1,000 ppm)	0.14 µS	25°C
#4 (5,000 ppm)	0.66 µS	25°C
#5 (10,000 ppm)	1.26 µS	25°C
Part 2: Conductance	of Unknown Solutions	(100 mL each)
Unknown Brine Solutions	Conductivity	Temperature (with thermometer)
Unknown #1	9.99 µS	25°C
Unknown #2	5.08 µS	25°C
Unknown #3	3.42 µS	25°C

Conclusions:

1. Using your data in part 1, calculate the molarity of each known brine solution. Show your calculations.

#1: __

#2: __

#3: __

#4: __

#5: __  

2.Based off your data in part 1 and the calculations found in Question #1^, calculate the concentration (M) of each unknown brine solution. Show your calculations and explain your reasoning.

Unknown #1: __

Unknown #2: __

Unknown #3: __

3. Using the information found in thr introduction, describe why the conductivity of whole milk is different than skim milk. Explain your reasoning.  

Answer 1

Question 1.

Molarity of solution#1: 100 ppm = 100 mg/L = (0.1 g/58.5 g.mol^-1)/L = 0.00171 mol/L = 0.00171 M

Molarity of solution#2: 500 ppm = 500 mg/L = (0.5 g/58.5 g.mol^-1)/L = 0.00855 mol/L = 0.00855 M

Molarity of solution#3: 1000 ppm = 1000 mg/L = (1 g/58.5 g.mol^-1)/L = 0.0171 mol/L = 0.0171 M

Molarity of solution#4: 5000 ppm = 5000 mg/L = (5 g/58.5 g.mol^-1)/L = 0.0855 mol/L = 0.0855 M

Molarity of solution#5: 10000 ppm = 10000 mg/L = (10 g/58.5 g.mol^-1)/L = 0.171 mol/L = 0.171 M