1) If a polymer is prepared from 0.89 g of acrylic acid, and 0.3 mL of 8.0 M NaO

Question

1) If a polymer is prepared from 0.89 g of acrylic acid, and 0.3 mL of 8.0 M NaOH, what will the mass of the "dry" polymer be after all of the water is removed? Assume all reactions go to completion

Hint : You may want to construct a BCA table to determine the outcome of the neutralization step

2) If a polymer with a dry mass of 1.9 has a mass of 153.2 after swelling in deionized water, what is the swelling ratio?

3) How much solid sodium chloride would you need to make 55.3 mL of a 3.53% (mass %) NaCl solution? Assume the density of water is 1.00 g/mL, and that the density of the solution is close to that of pure water

Explanation / Answer

1) No of moles of acrylic acid = 0.89/72.06 = 0.012 mole

No of moles of NaOH = 0.3/1000*8 = 0.0024 mole

H2C=CH-COOH + NaOH ----> H2C=CH-COONa + H2O

SO that, 1 mole acrylicacid = 1 mole NaOH

limiting reagent is NaOH

so that No of moles of polymer = 0.0024 mole

mass of H2C=CH-COONa = 0.0024*94 = 0.2256 grams

2) swelling ratio

1.9:153.2   =   1:80.63

3) 3.53% (mass %) NaCl solution: means 3.53grams of NaCl present in 100grams of solution.

so that as density = 1 g/cc

3.53grams of NaCl present in 100 ml of solution.

therefore,

mass of NaCl = 55.3*3.53/100 = 1.9521 grams

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