7.67 g of Sr(NO3)2 is dissolved in enough water to form 0.850 L. A 0.100 L sample is withdrawn ...?
2008-09-21 1:59 am
In the laboratory, 7.67 g of Sr(NO3)2 is dissolved in enough water to form 0.850 L. A 0.100 L sample is withdrawn from this stock solution and titrated with a 0.0360 M solution of Na2CrO4. What volume of Na2CrO4 solution is needed to precipitate all the Sr2+(aq) as SrCrO4?
回答 (1)
2008-09-21 10:53 am
✔ 最佳答案
Moles Sr(NO3)2 = 7.67 g / 211.6334 g/mol = 0.0362M = 0.0362 / 0.850 L =0.0426
Moles in 0.100 L = 0.00426
= moles Na2CrO4
V = 0.00426 / 0.0360 =0.118 L => 118 mL
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