The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is

Two aqueous solutions $S_{1}S\_1$ and $S_{2}S\_2$ are separated by a semi-permeable membrane. $S_{2}S\_2$ has lower vapour pressure than $S_{1}S\_1$ of a non-volatile solute, Then

When 3g of a non-volatile hydrocarbon containing 94.4% carbon is dissolved in 100g benzene, the vapour pressure of benzene is lowered from 9.953 kPa to 9.823 kPa. The molecular formula of the hydrocarbon is

200 mL of an aqueous solution of a protein contains 1.26 g. The osmotic pressure of this solution at 300 K is found to be $2.57\times {10}^{-3}2.57\backslash times10^\{-3\}$bar. Molar mass of protein will be

Which of the following 0.1M aqueous solution will have lowest freezing point?

0.004 M $N{a}_{2}S{O}_{4}Na\_2SO\_4$ is isotonic with 0.01M glucose. Degree of dissociation of $N{a}_{2}S{O}_{4}Na\_2SO\_4$ is

Freezing point of a solution containing 0.1g of $K_3[Fe(CN{)}_6]K\_3[Fe(CN)\_6]$(mol. wt. 329) in 100g of water ($K_f=1.86{\text{ K kg mol}}^{-1}K\_f=1.86\backslash text\{\; K\; kg\; mol\}^\{-1\}$) is

A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution.

The solutions A, B, C, and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M $BaC{l}_{2}BaCl\_2$ and 0.1 M $AlC{l}_{3}AlCl\_3$. which one of the following pairs is isotonic?

A solution containing 0.10 g of non-volatile solute X (molar mass = 100) in 200 g of benzene 0 depresses the freezing point of benzene by ${0.25}^{\circ }C0.25^\backslash circ\; C$ while 0.50 g of another non-volatile solute 0 Y in 100 g of benzene by ${0.25}^{\circ }C0.25^\backslash circ\; C$. What is the molar mass of Y?