A sample of common salt obtaied from sea water contain 39 g of NaCl and 1 g KCl. If it is dissolved in 100 g of water and then 90 g water is evaporated, what observation do you find at 0^(@)C? What is the maximum amount of pure NaCl that can be obtained by the above method ? (Solubilities of NaCl and KCl aer 40 g, 55 g to 100^(@)C and 35 g 28 g at 0^(@)C respectivety)

A sample of common salt obtaied from sea water contain 39 g of NaCl an

1.	A sample of common salt obtaied from sea water contain 39 g of NaCl and 1 g KCl. If it is dissolved in 100 g of water and then 90 g water is evaporated, what observation do you find at 0^(@)C? What is the maximum amount of pure NaCl that can be obtained by the above method ? (Solubilities of NaCl and KCl aer 40 g, 55 g to 100^(@)C and 35 g 28 g at 0^(@)C respectivety)
Answer» Solution :At `100^(@)C,` the solubilites NaCl and KCl are 40 and 55 respectively. Therefore entire sample dissolves in 100 g of wae. When 90 g of water is evaporated, 10 g of wate remains and since the solution to cooled to `0^(@)C` it can contain 3.5 g NACl and 2.8 g KCl.That means the entire KCl (1 gram) remains in solution which remains unsaturated wrt KCl.In 10G of watrer only 3.5 g remains in DISSOLVED state thereby making the solution saturated wrt NaCl. The rest of the NaCl that is 35.5 g gets precipitated out. This is the maximum amount of NaCl that can to obtained in pure FORM.