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Osmotic pressure of a 5% of a solution of cane sugar at 342K

Osmotic pressure of a 5% of a solution of cane sugar at 342K

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Grade:12th pass

2 Answers

Arun
25750 Points
4 years ago

 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

Vikas TU
14149 Points
4 years ago
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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