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Osmotic pressure of a 5% of a solution of cane sugar at 342K
5% means in 100 ml of solvent there is 5 g of solute, T=288 K,RMM=342g/mol, r=0.0821 lit atm K-1mol-1C is molarity, i.e no of moles of solute in a litre of solutionwe have pi=CRT=5/342*0.1 *0.0821 *288= 3.45 atm
5% means in 100 ml of solvent there is 5 g of solute, T=288 K,RMM=342g/mol, r=0.0821 lit atm K-1mol-1
C is molarity, i.e no of moles of solute in a litre of solution
we have pi=CRT
=5/342*0.1 *0.0821 *288= 3.45 atm
Dear student m = mol. mass of sucrose (C12H22O11) = 342 w = 5 g, V = 100 mL = 0.1 litre S = 0.082, T = (15 + 273) = 288 K Applying the equation PV = w/m ST, P = 5./342 × 1/0.1 × 0.082 × 288 = 3.453 atm
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