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Glucose is added to 1 litre water to such an extent that delTf/Kf becomes equal to 1/1000 the wt. of glucose added is: a.18g b.0.18g

Glucose is added to 1 litre water to such an extent that delTf/Kf becomes equal to 1/1000 the wt. of glucose added is:
a.18g         b.0.18g
 

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2 Answers

Amol Shegokar
13 Points
4 years ago
Depression in freezing point \Delta Tf = m . Kf
 
\therefore \Delta Tf/Kf = m
let ‘x’ g of glucose is added in 1 L of water (1kg of water as density is 1g/cc for water)
 m = molality = \frac{moles of solute}{kg of solvent} = \frac{x/180}{1}
mol. wt of glucose = 180 and moles = wt / mol wt.
\therefore \frac{x}{180} = \frac{1}{1000}
x = 0.18 g of glucose
 
Ritankar Das
13 Points
3 years ago
∆Tf =Kf × M
So, M = ∆Tf/Kf = 1/1000
M = moles of solute / volume of solution in litre
Volume of solution = 1litre 
So, moles of glucose = 1/1000 = 0.001
Molar mass of glucose = 180 g
Mass of glucose required = 180 × 0.001= 0.18 g

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