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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?
@ 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 H2O2- Calculate moles Sr(OH)2 remaining: 3.13 g Sr(OH)2 / 121.6 g/mol = 0.0257 moles Sr(OH)23- Calculate the ratio of moles H2O / moles Sr(OH)2: 0.207 / 0.0257 = 8.0so, here the value of x will turns out to be 8 . HOPE IT CLEARS NOW .. ALL THE BEST ..
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