Related Questions
Consider the following aqueous solutions and assume 100% ionisation in electrolytes
I. 0.1 m urea
II. 0.04 m
III. 0.05 m
IV. 0.005 m NaCl
The correct statement regarding the above solution is
Freezing point will be lowest for solution I
Freezing point will be highest for solution IV
Boiling point will be highest for solution IV
Vapour pressure will be highest for solution II
A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in of an aqueous solution. At an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is : (Assume density of solution is )
A solution of 4.5 g of a pure non-electrolyte in 100 g of water was found to freeze at . The molecular weight of the solute closest to
is isotonic with 0.01 M glucose.Degree of dissociation of is
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If for water is , the lowering in freezing point of the solution is
The freezing point depression of , is . If for water, , value of x will be
4
3
2
1
What is the amount of urea dissolved per litre if its aqueous solution is isotonic with 10% cane sugar solution? (mol.wt.of urea =60)
200 g/L
19.2 g/L
17.54 g/L
16.7 g/L
20g of glucose, 10g of urea, and 30g of sucrose are dissolved separately in 500ml of water each. Arrange these solutions in decreasing order of their osmotic pressure.
Glucose > Urea > Sucrose
Sucrose > Glucose > Urea
Urea > Glucose > Sucrose
Sucrose > Urea > Glucose
A solution containing 1.8 g of a compound (empirical formula ) in 40 g of water is observed to freeze at The molecular formula of the compound is
When a substance is distributed between two immiscible solvents and remains in the same state in the solvent I, while, dissociates in the solvent II. If the concentration of solute are in phase I and II respectively then :