Answer to Question #95151 in General Chemistry for Emily

Question #95151
0.895 g of MnI2 • x H2O gave 0.725 g of anhydrous compound after heating to remove the water. How many moles of water per mole of MnI2 are in the hydrate?
1
Expert's answer
2019-09-25T05:46:05-0400

In anhydrous compound H2O molecule comes out from the coordination cage

so in 0.895g of MnI2 0.17 g water as the weight of anhydrous molecule 0.725g

Molecular weight of water 18g

molecular weight of anhydrous MnI2 308.74 gm

So in 0.895g of hydrate 0.00234 mole of MnI2 present and 0.0094 mole of H2O present

So in 1 mole of MnI2 anhydrous contain 4 mole of water


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