Solutions Questions

Which of the following aqueous solutions will exhibit the highest boiling point?

Match List - I with List - IIList-I (Example)A. HumidityB. AlloysC. AmalgamsD. SmokeList-II (Type of Solution)I. Solid in solidII. Liquid in gasIII. Solid in gasIV. Liquid in solidChoose the correct answer from the options given below:

The freezing point depression constant for water is $1.86^{\circ }\text{C/m}$. If $5.00\text{g}$ ${\text{Na}}_2{\text{SO}}_4$ is dissolved in $45.0\text{g}$ ${\text{H}}_2\text{O}$, the freezing point is changed by $-3.82^{\circ }\text{C}$. Calculate the van't Hoff factor for ${\text{Na}}_2{\text{SO}}_4$: (Molar mass of ${\text{Na}}_2{\text{SO}}_4=142\text{g/mol}$)

The van't Hoff factor $i$ for a compound which undergoes dissociation in one solvent and association in another solvent is respectively:

Mole fraction of the solute in a $1.00$ molal aqueous solution is:

Concentrated aqueous sulphuric acid is $98\%{\text{ H}}_2{\text{SO}}_4$ by mass and has a density of $1.80{\text{ g mL}}^{-1}$. The volume of acid required to make one litre of $0.1{\text{ M H}}_2{\text{SO}}_4$ solution is:

$0.5\text{ molal}$ aqueous solution of a weak acid ($\text{HX}$) is $20\%$ ionised. If $K_f$ for water is $1.86{\text{ K kg mol}}^{-1}$, the lowering in freezing point of the solution is:

Which one of the following is incorrect for ideal solution?

The van't Hoff factor ($i$) for a dilute aqueous solution of the strong electrolyte barium hydroxide is:

Which of the following statements about the composition of the vapour over an ideal $1:1$ molar mixture of benzene and toluene is correct? Assume that the temperature is constant at $25^{\circ }\text{C}$. (Given vapour pressure data at $25^{\circ }\text{C}$: benzene $=12.8\text{kPa}$, toluene $=3.85\text{kPa}$)

At $100^{\circ }\text{C}$ the vapour pressure of a solution of $6.5\text{g}$ of a solute in $100\text{g}$ water is $732\text{mm}$. If $K_b=0.52$, the boiling point of this solution will be:

What is the mole fraction of the solute in a $1.00m$ aqueous solution?

Of the following $0.10\text{m}$ aqueous solutions, which one will exhibit the largest freezing point depression?

$p_A$ and $p_B$ are the vapour pressure of pure liquid components $A$ and $B$, respectively, of an ideal binary solution. If $x_A$ represents the mole fraction of component $A$, the total pressure of the solution will be:

Vapour pressure of chloroform (${\text{CHCl}}_3$) and dichloromethane (${\text{CH}}_2{\text{Cl}}_2$) at $25^{\circ }\text{C}$ are $200\text{mm Hg}$ and $41.5\text{mm Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5\text{g}$ of ${\text{CHCl}}_3$ and $40\text{g}$ of ${\text{CH}}_2{\text{Cl}}_2$ at the same temperature will be: (Molecular mass of ${\text{CHCl}}_3=119.5\text{u}$ and molecular mass of ${\text{CH}}_2{\text{Cl}}_2=85\text{u}$)

$200\text{mL}$ of an aqueous solution of a protein contains $1.26\text{g}$. The osmotic pressure at $300\text{K}$ is $2.57\times 10^{-3}\text{bar}$. The molar mass of protein is ($R=0.083\text{L}\text{bar}{\text{mol}}^{-1}{\text{K}}^{-1}$):

$0.1\text{molal}$ weak acid is $30\%$ ionized. $K_f=1.86$. Freezing point is:

A solution of sucrose (molar mass $=342{\text{g mol}}^{-1}$) has been prepared by dissolving $68.5\text{g}$ of sucrose in $1000\text{g}$ of water. The freezing point of the solution obtained will be ($K_f$ for water $=1.86{\text{K kg mol}}^{-1}$):

An aqueous solution is $1.00\text{ molal}$ in $\text{KI}$. Which change will cause the vapour pressure of the solution to increase?

What is the $[{\text{OH}}^{-}]$ in the final solution prepared by mixing $20.0\text{mL}$ of $0.050\text{M HCl}$ with $30.0\text{mL}$ of $0.10{\text{M Ba(OH)}}_2$?

A $0.0020\text{m}$ aqueous solution of an ionic compound ${\text{Co(NH}}_3{)}_5({\text{NO}}_2)\text{Cl}$ freezes at $-0.00732^{\circ }\text{C}$. Number of moles of ions which $1\text{mol}$ of ionic compound produces on being dissolved in water will be ($k_f=1.86^{\circ }\text{C/m}$):

During osmosis, the flow of water through a semi-permeable membrane is :

A solution of acetone in ethanol :

$1.00\text{g}$ of non-electrolyte solute (molar mass $250{\text{g mol}}^{-1}$) was dissolved in $51.2\text{g}$ of benzene. If the freezing point depression constant, $K_f$ of benzene is $5.12{\text{K kg mol}}^{-1}$, the freezing point of benzene will be lowered by :

A solution containing $10{\text{g per dm}}^3$ of urea (molecular mass = $60{\text{g mol}}^{-1}$) is isotonic with a $5\%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.Reason : Water is polar and benzene is non-polar.