Question

Prelab (10 pts.) Serial dilutions: see the information on P9 from Experiment 9: AMD for dilution calculations. 1. Calculate the volume of 0.05 M HCl needed to make 30.0 mL of a 0.025 M HCI solution (2 pts.). a. b. Calculate the volume of 0.025 M HCl needed to make 30.0 mL of a 0.0125 M HCI solution (2 pts.). 2. Use the information on electrolyte vs. nonelectrolytes found on page 328 in your text to answer the following questions (1 pt. each, 6 pts. total). A. Which of the following is a strong electrolyte? C. Which of the following is a nonelectrolyte? a. sodium chloride a. sodium chloride b. ethanol (CH3CH2OH) c. acetic acid (CH3COOH or HC,H3O2) b. ethanol (CH3CH2OH) c. acetic acid (CH3COOH or HC2H3O2) D. Which of the following is likely to have the highest conductivity? B. Which of the following is a weak electrolyte? a. Deionized water (DI) b. Tap water c. ethanol (CH3CH2OH) a. sodium chloride b. ethanol (CH 3CH2OH) c. acetic acid (CH3COOH or HC H3O2)

Answer 1

Q-1 Answer :

We know the relation between strength and volume of a solution on dilution from a stock solution -

V₁ × S₁ = V₂ × S₂     ……………….. (1)

Where, V₁ = volume of stock solution, S₁ = strength of stock solution, V₂ = volume of desired solution, S₂ = strength of desired solution

(a) V₁ = ?, S₁ = 0.05 M, V₂ = 30 mL, S₂ = 0.025 M

Putting values of above three in equation (1) we get,

V₁× 0.05 M = 30 mL × 0.025 M

V₁ = (30 mL × 0.025 M) / (0.05 M)

∴ V₁ = 15 mL

Therefore, 15 mL is needed to be added.

(b) V₁ = ?, S₁ = 0.025 M, V₂ = 30 mL, S₂ = 0.0125 M

Putting values of above three in equation (1) we get,

V₁× 0.025 M = 30 mL × 0.0125 M

V₁ = (30 mL × 0.0125 M) / (0.025 M)

∴ V₁ = 15 mL

Therefore, 15 mL is needed to be added.

Q-2. Answer:

(A) (a) sodium chloride (NaCl)

(B) (c) acetic acid (CH₃COOH or HC₂H₃O₂)

(C) (b) ethanol (CH₃CH₂OH)

(D) (b) Tap water