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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.
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Solution
Given: WA (of urea) = 15 g
Molar mass of urea = 60 g mol−1
Molar mass of glucose = 180 g mol−1
To find: WB (of glucose) = ?
Formula: `"Moles of urea" = "wt. of urea"/"molar mass"`
= `15/60`
= 0.25
For an isotonic solution,
π = CRT
(where, C is concentration, R is a constant, and T is temperature)
`π_"urea" = π_"glucose"`
`C_"urea" RT = C_"glucose" RT` ...(RT is common)
0.25 = n/1L ...(Taking C = n/V)
n glucose = 0.25
`"Moles of glucose" = "wt. of glucose"/"molar mass"`
0.25 = `W_B/180`
WB = 0.25 × 180
= 45 g
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