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A crystalline salt on being rendered anhydrous loses 45.6 % of its weight. The percentage composition of anhydrous salt is : `Al = 10.5 % , K = 15.1 % , S = 24.96 %, O = 49.92 %` Find the simplest formula of the anhydrous and crystalline salt. |
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Answer» Correct Answer - Anhydrous salt : `Kal_(2)S_(2)O_(8)` ; Hydrated salt : `KAlS_(2)O_(8).12H_(2)O` Step I. Formula of anhydrous salt `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Al",10.5,27,(10.50)/(27)=0.39,(0.39)/(0.38)=1.03,1),("K",15.1,39,(15.1)/(39)=0.38,(0.38)/(0.38)=1.00,1),("S",24.96,32,(24.96)/(32)=0.78,(0.78)/(0.38)=2.05,2),("O",49.92,16,(49.92)/(16)=3.12,(3.12)/(0.380)=8.21,8):}` Formula of anhydrous salt `= AlKS_(2)O_(8)` Empirical formula mass `= 27 + 39 + 2 xx 32 + 8 xx 16 = 258 u` Step II. Formula of crystalline salt Let the weight of hydrated salt = 100 u , Weight of water lost = 45.6 u Weight of anhydrous salt `= 100 - 45.6 u = 54.4 u` If the weight the anhydrous salt is 54.4 u, that of water = 45.6 u If the weight of anhydrous salt is 258 u, that of water `= ((45.6u))/((54.4u))xx(258u)=216.3u` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=((216.3u))/((18u))=12` (approx) `:.` The formula of crystalline salt `= KAlS_(2)O_(8).12H_(2)O`. |
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