3. Determine the % by mass of a solutio, when 12.5 grams of sodium chloride is d

Question

3. Determine the % by mass of a solutio, when 12.5 grams of sodium chloride is dissolved into 22.5 grams of water (225 ml).

4. What mass of barium chloride is neededto prepare 250 ml solution that has a concentration of 0.25 M?

5. What is the molarity of a solution when 54.5 grams of sodium sulfate is dissolved in water to make 500 ml of solution?

6. Wat is the molarity when 4.5 grams of sodim chloride is dissolved in 100 ml of water? (density of water=1.0 g/ml)

7. How many grams of sodium carbonate will react with25.5 ml of 0.300 M hydrochloric acid?

Na2CO3 + HCl > NaCl + H20 + CO2 (unbalanced)

a. Determine the number of moles (mmoles) hydrochloric acid used.

b. Determine the number of moles (mmoles) of sodium carbonate reacted.

c. Determine the mass (grams) of sodium carbonate reacted.

8. How many grams of aluminum hyfroxide will react with 75.5 ml of 0.250 M sulfuric acid?

Al(OH)3 + H2SO4 > Al2(SO4)3 + H20 (unbalanced)

a. Detrmine the number of mole (mmoles) sulfuric acid used.

b. Determine the number of moles (mmoles) Aluminum hydroxide reacted.

c. Detrmine the mass (grams) of aluminum hydroxide reacted.

9. Consider the following solutions. Put in order according to the boiling points from highest point to lowest. (Hint: Colligative properties. Kb=0.52 C/m for water)

a. 0.1 M c6h12o6 b. 0.1 M CaCl2 c. 0.1 M NaCl

Explanation / Answer

3) % by mass of NaCl in solution = [12.5 g / (12.5+22.5) g]*100 = 35.7 %

4) Weight of BaCl2 = (Molarity * Volume) *Molar mass = 0.25 mol/L * 0.25 L * 208.23 g/mol = 13.0 g

5) Molarity = (mass/Molar mass) / V (in L) = (54.5 g /142.04 g/mol) / 0.5 L = 0.77 mol/L

6) Molarity = (4.5 /58.45) mol / 0.1 L = 0.77 M

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