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Grade: 12

                        

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 _____

4 years ago

Answers : (2)

Vikas TU
12119 Points
							
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.
4 years ago
Krish Gupta
askIITians Faculty
76 Points
							
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.
3 months ago
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