A 6.85 sample oy the hydrate Sr(OH)².xH²O is dried in an even to give 3.13 g of anhydrous Sr(OH)².What is the value of x?

Question

Umakant biswal · Answer

@ suman
below i am giving you the approach to solve the questions . kindly follow that
1- Calculate the mass and moles of H2O driven off:
6.85 g - 3.13 g = 3.72 g H2O / 18.0 g/mol = 0.207 moles H2O
2- Calculate moles Sr(OH)2 remaining:
3.13 g Sr(OH)2 / 121.6 g/mol = 0.0257 moles Sr(OH)2
3- Calculate the ratio of moles H2O / moles Sr(OH)2:
0.207 / 0.0257 = 8.0
so, here the value of x will turns out to be 8 .
HOPE IT CLEARS NOW ..
ALL THE BEST ..

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A 6.85 sample oy the hydrate Sr(OH)².xH²O is dried in an even to give 3.13 g of anhydrous Sr(OH)².What is the value of x?

A 6.85 sample oy the hydrate Sr(OH)².xH²O is dried in an even to give 3.13 g of anhydrous Sr(OH)².What is the value of x?

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