26.7 g of a compound with the formula PtCl_2 | vedclass

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(B) The compound $PtCl_2Br_2 \cdot 6H_2O$ can be represented as $[Pt(H_2O)_3Cl_2Br]Br \cdot 3H_2O$ or $[Pt(H_2O)_3Br_2Cl]Cl \cdot 3H_2O$ to satisfy th

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$2.70$

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(B) The compound $PtCl_2Br_2 \cdot 6H_2O$ can be represented as $[Pt(H_2O)_3Cl_2Br]Br \cdot 3H_2O$ or $[Pt(H_2O)_3Br_2Cl]Cl \cdot 3H_2O$ to satisfy the coordination number of $6$ and the existence of $3$ stereoisomers (geometrical isomers).Number of moles of the compound $= \frac{26.7 \ g}{534 \ g \ mol^{-1}} = 0.05 \ mol$.In the formula,there are $3$ molecules of water of crystallization.Moles of water of crystallization $= 0.05 	imes 3 = 0.15 \ mol$.Concentrated $H_2SO_4$ acts as a dehydrating agent and absorbs the water of crystallization.Mass of water lost $= 0.15 \ mol 	imes 18 \ g \ mol^{-1} = 2.7 \ g$.

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