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A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the anhydrous salt is : {:("Aluminium"=10.50%,,"Potassium"=15.1%,),("Sulphur"=24.96%,,"Oxygen"=49.92%.):} Find the simplest formula of the anhydrous and crystalline salt.Find the simplest formula of the anhydrous and crystalline salt. |
Answer» Solution :Step 1. To calculate the empirical formula of the anhydrous salt  Thus, the empirical formula of the anhydrous salt is K Al `S_(2)O_(8)` Step 2. To calcualte the empirical formula mass of the anhydrous salt. Empirical formula mass of the anhydrous salt `(KAlS_(2)O_(8))=1xx39.0+1xx27.0+2xx32.0+8xx16.0` `=258.0u.` Step 3. To calculate the empirical formula mass of the hydrated salt. Let the empirical formula mass of the hydrated salt = 100.0 u. Loss of weight due to dehydration = `45.6%` `therefore` Empirical formula mass of the anhydrous salt `=100-45.6=54.5u.` Now, if the empirical formula mass of the anhydrous salt is 54.4, then that of hydrated salt = 100 `therefore `If the empirical formula mass of the anhydrous salt is 258,, that of hydrated salt `=(100)/(54.4)xx258=474.3u.` Step 4. To calculate the number of molecules of WATER in the hydrate salt. Total loss in mass due to dehydration =`474.3-258.0=216.3u` Loss in mass due to one MOLECULE of water= 18.0 u `therefore"No. of molecules of water in the hydrated SAMPLE"=(216.3)/(18)=12` Step 5. To calcualte the empirical formula of the hydrated salt. Empirical formula of the anhydrous salt `=KAlS_(2)O_(8)` No. of molecules of water of crystallization = 12 `"Empirical formula of the anhydrated salt "=KAlS_(2)O_(8).12H_(2)O` |
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