Relative Lowering of Vapour Pressure

Hoar frost is -

The mass of a non-volatile, non-electrolyte solute (molar mass $= 50 g mo{l}^{-1}$ ) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75 %, is:

A solution $\mathrm{X}$ contains $12 \mathrm{g}$ of urea per litre of water. Another solution $\mathrm{Y}$ contains $68.4 \mathrm{g}$ of cane sugar per litre of water at the same temperature. What is the ratio of lowering of vapour pressure in solutions of $\mathrm{X}$ and $\mathrm{Y}$?

$\left(\mathrm{N}=14 \mathrm{u}, \mathrm{C}=12 \mathrm{u}, \mathrm{O}=16 \mathrm{u}, \mathrm{H}=1 \mathrm{u}\right)$

The vapour pressure of an aqueous solution of cane sugar (molar weight=$342$) is $756 \mathrm{mm}$ at $100°\mathrm{C}$. How many grams of sugar are present per $1000 \mathrm{g}$ of water? Report the answer in the nearest integer

What is the vapour pressure of a solution containing $15.6 \mathrm{g}$ of water and $1.68 \mathrm{g}$ of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ at $100° \mathrm{C}$?

Give your answer by rounding off to the nearest integer value.Molecular weight of sucrose is $342$.

The pressure of the water vapour of a solution containing a non-volatile solute is $2\%$ below that of the vapour of pure water. Determine the molality of the solution.

Round off your answer to nearest integer value.

An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004$ bar at the normal boiling point of the solvent. What is the molar mass of the solute? (Write the answer without units)

Two beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of naphthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.

The molar mass of solute in solution B is closest to:

Relative vapour pressure lowering depends only on.

Vapour pressure of solution of a non volatile solute is always

Lowering of the vapour pressure of the solution:

Find the value of mole fraction of the solute when vapour pressure of glucose is $750 \mathrm{mm} \mathrm{Hg}$ at $373 \mathrm{K}$ in dilute solution.

Find the correct relation between ${\mathrm{pK}}_{\mathrm{a}1} \mathrm{and} {\mathrm{pK}}_{\mathrm{a}2}$ of acids HA₁ and HA₂ (assuming they are taken in same concentration), if lowering of vapour pressure $\Delta \mathrm{P}_{1}$ of an aqueous solution of acid HA₁ and $\Delta \mathrm{P}_{2}$ of an aqueous solution of acid HA₂, are related as $∆{\mathrm{P}}_1>∆{\mathrm{P}}_2$.
 

An ideal solution of two pure liquids $\mathrm{A}$ and $\mathrm{B}$ are having the vapour pressure of $100$ and $400 \mathrm{torr}$ respectively at the temperature $\mathrm{T}$. The liquid solution of $\mathrm{A}$ and $\mathrm{B}$ is made up of $1 \mathrm{mole}$ of each $\mathrm{A}$ and $\mathrm{B}$. Then, find the pressure when $1 \mathrm{mole}$ of mixture has been vaporised.

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

A solution of 2.0 g of sulphur in 100 g of CS₂ has a vapour pressure of 844.9 torr. At 48^oC, the vapour pressure of pure CS₂ is 850 torr. Calculate the atomicity of sulphur molecule.

If two beaker $\mathrm{A}$ and $\mathrm{B}$ are present in a closed vessel. Beaker $\mathrm{A}$ contains$152.4 \mathrm{g}$ aqueous solution of urea, containing$12 \mathrm{g}$ of urea. Beaker $\mathrm{B}$ contains $196.2 \mathrm{g}$ glucose solution, containing $18 \mathrm{g}$ of glucose. Both solutions allowed to attain the equilibrium. Calculate the Weight% of glucose in its solution at equilibrium.

The vapor pressure of pure  water $\left({\mathrm{P}}^{\mathrm{o}}\right)$ is $17.5\text{mm}$ of $\mathrm{Hg}$ at $20°\text{C}$. Calculate the vapor pressure of the resulting solution $\left({\mathrm{P}}^{\mathrm{s}}\right)$  when $18\text{g}$ of glucose is added to $178.2\text{g}$ of water?

Assertion : When dry air is passed continuously through an aqueous solution of a non-volatile solute, vapour pressure of solution decreases gradually.

Reason : Some air is absorbed in the solution.

The vapour pressure of water at $20°\mathrm{C}$ is $17.5\text{ mm of Hg}.$ If $18 \mathrm{g}$ of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ is added to $178.2 \mathrm{g}$ of water at ${20}^{\circ }\mathrm{C}$, what will be the vapour pressure of the resulting solution?