Question

Calculate the percent error as a result of using 125mL of 25° C water, compared with the correcr yield using 20mL of 4° C water.

124 SYNT0341: Synthesis of Strontium Iodate Monohydrate 3. Assume that, in the experiment described in Post-Laboratory Question 2, 125 mL of 25°C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (at 4 °C). The solubility of barium iodate monohydrate in 25°C water is 0.028 g per 100 mL of water; in 4 °C water, it is 0.010 g per 100 mL of water. You H2O = 0.010g/100ml (1) What mass of product would you expect to isolate? vowme H2O @ 4.c 20mL 10.010er bomL) 0.oong 100mL mans is procurd isolated will se o oudy Comes as it is waoned i dissolnod in 4°C sont to. (2) Calculate the percent error as a result of using 125 mL of 25 °C water, compared with the correct yield using 20 mL of 4 °C water. n on NO od )

Answer 1

(1)

Solubility of barium iodate monohydrate in water at 25ºC = 0.028 g per 100 mL.

Therefore, mass of barium iodate monohydrate isolated from 125 mL water = (125 mL)*(0.028 g)/(100 mL)

= 0.035 g (ans).

(2)

Solubility of barium iodate monohydrate in water at 4ºC = 0.010 g per 100 mL.

Therefore, mass of barium iodate monohydrate expected from 20 mL of water at 4ºC = (20 mL)*(0.010 g)/(100 mL)

= 0.002 g.

Percent error = [(mass isolated – mass expected)/(mass expected)]*100 %

= [(0.035 g – 0.002 g)/(0.002 g)]*100 %

= [(0.033 g)/(0.002 g])*100 %

= 1650 %