# Chemistry

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A student synthesized 6.895 g of barium iodate monohydrate by adding 30.00 mL of 5.912 x 10-1 M sodium iodate, NAIO3. 125 mL of distilled water as 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 .028 g per 100 mL of water; in 4* C water, it is .010 g per 100 mL of water.

1) What mass of product would you expect to isolate?

2) Calculate the percent error as a result of using 125 mL of 25* C, compared with the correct yield using 20 mL of 4* C water.

• Chemistry -

I can't make sense of the problem, partly because how it is stated and partly because of the sentence structure. I don't know if the 6.895 is the theoretical yield OR if it is the actual yield or it is the actual yield if the chilled water is procedure is followed. Also, it isn't clear if the volume used is 125 or 125+35= 160 mL. I suspect actual yield IF the chilled procedure was followed and 125 mL volume so the following answers are based on those assumptions.
Solubility if the student followed directions for 20 mL chilled water is
0.01g/100mL x (20/100) = 0.002 g Ba(IO3)2 lost due to solubility.

Solubility if the student used 125 mL water at 25 C is
0.028g/100mL x 125/100 = 0.035 g lost due to solubility Ba(IO3)2.

So the student would have collected 6.895g if the chilled water procedure was used (with the assumptions made above). Since 0.035 g was lost, the student actually collected 6.895 - 0.035 = 6.860 g.

%error = (0.035/6.895)*100 = ?

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