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Water of crystallisation calculations. A 0.970 g sample of a hydrated copper (II) salt, CuX 2 .6H 2 O was heated to give 0.538 g of the anhydrous salt. What is X?

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## Water of crystallisation calculations

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**Water of crystallisation calculations**A 0.970 g sample of a hydrated copper (II) salt, CuX2.6H2O was heated to give 0.538 g of the anhydrous salt. What is X? A 46.1 g sample of hydrated potassium dichromate, K2Cr2O7.xH2O was heated to give 35.3 g of the anhydrous salt. What is the value of x?**A 0.970 g sample of a hydrated copper (II) salt, CuX2.6H2O**was heated to give 0.538 g of the anhydrous salt. What is X? Mass of H2O = mass of hydrated salt – mass of anhydrous salt = 0.970 – 0.538 = 0.432 g Moles of H2O = 0.432/18.0 = 0.024 moles Ratio CuX2 : H2O 1 : 6 So moles CuX2 = 0.024/6 = 0.004 Mr = mass/moles = 0.538/0.004 = 134.5 Mr of ‘X’ = (134.5-63.5)/2 = 35.5 X = Cl**A 46.1 g sample of hydrated potassium dichromate,**K2Cr2O7.xH2O was heated to give 35.3 g of the anhydrous salt. What is the value of x? Moles of anhydrous salt = 35.3/294.2 = 0.120 moles Mass of H2O = 46.1 – 35.3 = 10.8 g Moles of H2O = 10.8/18.0 = 0.600 moles x = 0.600/0.120 = 5**The salt XSO4.5H2O was heated and 0.5 moles of water was**lost. It left behind an anhydrous mass of 15.96g. What is X?**The salt XSO4.5H2O was heated and 0.5 moles of water was**lost. It left behind an anhydrous mass of 15.96g. What is X? n XSO4 = 0.1 Mass XSO4 =15.96g Mr =mass/n Mr =15.96/0.1=159.6 Ar X =159.6-(32.1+64)=63.5 Therefore X = Cu

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