Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution. - Chemistry

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Question

A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution.

Solution 1

For isotonic solution

π1 = π2

C1 = C2 (at same temperature)

or n1 = n2 (is same volume)

`15/60=x/180`

x = 45 g, mass of glucose per liter of solution.

Solution 2

Osmotic pressure π = CRT =`(n2)/VRT`

where n is the number of moles of solute present in volume V of solvent.

It is given that the solution of urea is isotonic with the solution of glucose, thus

`(n_`

`=>(n_`

`=>(w_`

`=>w_`

Hence, the mass of glucose present in one litre of its solution = 45 g

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Solution A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution. Concept: Colligative Properties and Determination of Molar Mass - Osmosis and Osmotic Pressure.
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