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If 18g of glucose `(C_(6)H_(12)O_(6))` is present in 1018 g of an aqueous solution of glucose, it is said to beA. 1 molalB. 1.1 molalC. 0.5 molalD. 0.1 molal

Answer» Correct Answer - D
18 g glucose `=18/180 ` mol =0.1 mol.
As it is present in (1018-18) = 1000 g water, the solution is 0.1 molal.


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