An aqueous solution of urea has freezing point of -0.604^(@)C & At 27^(@)C . The osmatic pressure of the same solution is ____________atm . (assume molality and molarity are same)
The depression in feezing point is 0.93^(@)C then the osmotic pressure of aqueous solution of the given non-electrolyte at 27^(@)C is
18 g glucose and 6 g urea are dissolved in 1L of solution at 27^@C , the osmotic pressure in atm of the solution will be
300 mL of an aqueous solution of a protein contains 2.52 g of the protein. If osmotic pressure of such a solution at 300 K is 5.04xx10^(-3) bar, the molar mass of the protein in g mol^(-1) is
One liter aqueous solutions contains 2 xx 10^(-2) kg glucose at 25^@C . Calculate the osmotic pressure of the solution.
The boiling point of an aqueous solution is found to be 100.28^(@)C . Then the freezing point of the same solution is -X^(0)C . What is the value of 'X' ?
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
