During osmosis, flow of water through a semi-permeable membrane 
During osmosis, flow of water through a semi-permeable membrane is from solution having lower concentration only.
A solution containing 10 g per dm 3 of urea (molecular mass = 60 g mol–1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this nonvolatile solute is 
1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by 
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution is
As ΔTf = iKfm
Total no. of moles = 1 – 0.20 + 0.20 + 0.20 = 1 + 0.20 = 1.2
∴ ΔTf = 1.2 × 1.86 × 0.5 = 1.1160 ≈ 1.12 K
A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)
ΔTf = 0 – (0.00732°) = 0.00732
ΔTf = i × Kf × m
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase? 
When the aqueous solution of one molal KI is diluted with water, concentration decreases, therefore the vapour pressure of the resulting solution increases.
A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ( f for water = 1.86 K kg mol–1). 
T = – 0.372°C
2 5. 3 g of sodium carbonate, Na2 CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, Na+ and carbonate ions, CO32– are respectively (Molar mass of Na2CO3 = 106 g mol–1) 
[Na+] = 2 × 0.955 = 1.91 M
= 0.955 M
The freezing point depression constant for water is – 1.86ºC m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by – 3.82ºC. Calculate the van’t Hoff factor for Na2SO4. 
Given Kf = – 1.86º cm–1, mass of solute = 5.00 g, mass of solvent = 45.0 g
∴ i = 2.63 (Molecular mass of Na2SO4 = 142)
The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively : 
If compound dissociates in solvent i > 1 and on association i < 1.
Mole fraction of the solute in a 1.00 molal aqueous solution is : 
1 molal solution means 1 mole of solute dissolved in 1000 gm solvent.
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]
Given α = 30% i.e., 0.3
i = 1.3
ΔTf = 1.3 × 1.86 × 0.1 = 0.2418
Tf = 0 – 0.2418 = – 0.2418 °C
200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10–3 bar.The molar mass of protein will be (R = 0.083 L bar mol–1 K–1) [2011 M]
(π) = CRT Osmotic pressure
= 2.57 10 3
Molecular mass = 61038 g
PA and PB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If XA represents the mole fraction of component A, the total pressure of the solution will be. 
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25ºC are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be : (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u). [2012 M]
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3 ? The concentrated acid is 70% HNO3 [NEET 2013]
wt. of 70% acid
Which condition is not satisfied by an ideal solution? [NEET Kar. 2013]
An ideal solution is that solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations. For an ideal solution.