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if 18 gram of glucose is present in 1000 gram of an aqueous solution of glucose it is said to be how much molal

if 18 gram of glucose is present in 1000 gram of an aqueous solution of glucose it is said to be how much molal

Grade:12

2 Answers

Norman Manchu
151 Points
6 years ago
Molality is defined as n number of moles of solute dissolved in per kilograms of the solvent So it is needed that we find the weight of the solvent which not being mentioned has to be assumed water thus called aqueous!m=n(glucose)|Weight in kgs(water)=n(glucose)x1000|Weight of water in grams. =(18÷180)x1000|(1000-18) = 0.10molal
ankit singh
askIITians Faculty 614 Points
3 years ago
18 g of glucose → 18/180 mole or 0.1 mole of glucose. Here, 0.1 mole of glucose present in 1000 g of solvent. On the basis of known definition i.e. molality, so, the concentration of the glucose in the solvent would be 0.1 molal or simply 0.1 m.

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