find the amount of Ammonium chloride that will separate out when 155g of its solution at 333k is cooled to 293 k
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Solubility of ammonium chloride at 293 K = 37 g
So, amount of ammonium chloride that will separate out when 155g of its solution at 333K is cooled to 293K = 155 - 37
= 118 g
So, amount of ammonium chloride that will separate out when 155g of its solution at 333K is cooled to 293K = 155 - 37
= 118 g
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