Answer to Question #95151 in General Chemistry for Emily

Question

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?

Answer 1

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