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osmotic pressure of 40% (mass/vol.) urea solution is 1.64 atm and that 3.42%(mass/vol.) cane-sugar is 2.46 atm . when equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is _____

krishna priya , 8 Years ago
Grade 12
anser 2 Answers
Vikas TU

Last Activity: 8 Years ago

let M1 be the molarity of urea soln. and M2 be the molarity for Cane -sugar soln. respectively.
Of equal volumes would be mixed then,
M1V + M2V = M(2V)
M = (M1+M2)/2 would be the new concentration.
 
From first soln.
1.64 =M1ST
or 
M1 = 1.64/ST
similarly
M2 = 2.46/ST
Put it in M.
then apply for mixed soln.
O.P = 1*M*S*T
S*T would be cancelled out.

Krish Gupta

Last Activity: 4 Years ago

Yes,
let M1 be the molarity of urea soln. and M2 be the molarity for Cane -sugar soln. respectively.
Of equal volumes would be mixed then,
M1V + M2V = M(2V)
M = (M1+M2)/2 would be the new concentration.
 
From first soln.
1.64 =M1ST
or 
M1 = 1.64/ST
similarly
M2 = 2.46/ST
Put it in M.
then apply for mixed soln.
O.P = 1*M*S*T
S*T would be cancelled out.

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