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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 _____
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 =M1STor M1 = 1.64/STsimilarlyM2 = 2.46/STPut it in M.then apply for mixed soln.O.P = 1*M*S*TS*T would be cancelled out.
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 =M1STor M1 = 1.64/STsimilarlyM2 = 2.46/STPut it in M.then apply for mixed soln.O.P = 1*M*S*TS*T would be cancelled out.
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