CHEM 1151 L Lab 14 - Lab 14 PDF

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Lab 14...

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Pre-Lab Study Questions

14

1. Why does an oil-and-vinegar salad dressing have two separate layers?

Oil and vinegar will nit dissolve because vinegar has a higher density

2. What is meant by the mass percent (m/m) concentration of a solution?

Expresses the grams of absolute in the grand of a solution 3. What is molarity? A way to express concentrations which more specifically is the number of miles per liter of the solution

4. Why are some electrolytes considered strong, whereas others are considered weak? It is determines by how much a solute disassociate any solution

5. A solution is prepared with 3.26 g KCl and water to make 25.0 mL of KCl. a. What is the % (m/v) of the KCl solution? 13.4%

Questions and Problems Q1 Why are some solutes soluble in water, but others are soluble in cyclohexane? Polar molecules can interact with the partial positive and partially negative part of polar solvents such water forming dipole dipole interactions can dissolve in water

Q2 For the three solutes tested in B, write an equation for their dissolution in water: HCL+H20H30^+ +CL^-

HCL(g)

NH4OH+H20NH3+H20

NH4OH(aq) C6H12O6(s)

C2H5OH

Q3 Classify the solutes in each of the following equations as a weak electrolyte, a strong electrolyte, or a nonelectrolyte in water: a. XY2 (s)    X 2  (aq)  2Y – (aq)

STRONG

  H  (aq )  X– (aq ) b. HX(g )   

WEAK NON

c. XYZ( s)   XYZ( aq) d. YOH(s)   Y  (aq)  OH – (aq )

STRONG

C. Electrolytes in Body Fluids 1. Type of IV Solution

5% Isolyte P (mEq/L)

Plasnalyte (mEq/L)

Na(23),K(20), Mg(3)

Na(140), K(5), Mg(3)

Cl(109), Lactate (28)

Cl(29),Acetate (23), Phosphate(HPO4)=

Cl(98),Acetate(27), Gluconate(23)

4. Total Charge of Cations (+)

137

46

148

5. Total Charge of Anions ()

137

55

148

2. Cations

3. Anions

Lactated Ringer’s (mEq/L) Ca(3), K(4), Na(130)

6. Sum of the Charges

274

101

296

Q4 What would be the overall charge in any IV solution? Why? The overall charge should be 0 because the charge of the cations cancels out the anions to become neutral Q5 A 15.0-mL sample of NaCl solution has a mass of 15.78 g. After the NaCl solution is evaporated to dryness, the dry salt residue has a mass of 3.26 g. Calculate the following concentrations for the NaCl solution.

a. % (m/m) 20.1% m/m

b. % (m/v) 21.7% m/v

c. molarity (M) 3.72 M

Q6 How many grams of KI are in 25.0 mL of a 3.0 % (m/v) KI solution? (Use % as a conversion factor) .75 g

Q7 How many milliliters of a 2.5 M MgCl 2 solution contain 17.5 g MgCl 2? (Use M as a conversion factor) 73.52 ml...