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The mass percent of water in a hydrate of mncl2 is 36.41 percent what is the fomular of the hydrate?
Wiki User
∙ 11y ago
36.41% = .3641 g of H2O
63.59% = .6359 g of MnCl2
.6359g MnCl2 x (1molm MnCl2/125.84g MnCl2)=.0051 mol MnCl2
.3641g H2O x (1 mol H2O/ 18.02g H2O) = .0202 mol H2O
.0051 mol MnCl2 / .0051 = 1mol MnCl2
.0202 mol H2O /.0051 = 3.96 = 4 mol H2O
Formula is: MnCl2 4H2O Manganese (II) chloride tetrahydrate
Wiki User
∙ 11y ago
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high
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Suppose your sample contain a volatile impurity.would your determination of percent of water in hydrate be too high or low explain your answer?
high
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