A) 0.01m
B) 0.617m
C) 0.668m
D) 1.623m
Correct Answer: B
Solution :
Mass of solution = 100 g Mass of glucose = 10 g, Mass of water = 90 g No. of moles of glucose = \[\frac{10}{180}\]= 0.0555 mol No. of moles of water = \[\frac{90}{18}\] =5 mol Molality=\[\frac{No.\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kg}=\frac{0.0555\,mol}{0.090\,kg}\] = 0.617mYou need to login to perform this action.
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