Calculate the temperature at which a solution containing 54g of glucose, c6h12o6, in 250g of wate will freeze
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molality of the glucose =no of mole of glucose in 1 kg of water
no of mole of glucose =weight of the glucse/molecular weight of the glucose =54/180 =.3
we have ,delta Tf=Kfm
delta Tf= (1.86*.3)/.250
=2.232
temperature at which it will freeze=273 -2.232 K =270.768 K
no of mole of glucose =weight of the glucse/molecular weight of the glucose =54/180 =.3
we have ,delta Tf=Kfm
delta Tf= (1.86*.3)/.250
=2.232
temperature at which it will freeze=273 -2.232 K =270.768 K
Answered by
3
Explanation:
answer is -2.23 degree Celsius
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