Chemistry

Solutions and Vapour Pressure

60 Questions

Solutions and vapour pressure questions explore boiling points, colligative properties, and osmotic pressure. These concepts are regularly tested in competitive chemistry exams. Solving these requires a solid grasp of solute and solvent interactions.

colligative propertiesvapour pressureboiling and melting pointsosmotic pressure

Solutions and Vapour Pressure Questions

Multiple choice chemistry electrolysis electrolytes and non-electrolytes introduction to electrolysis chemical reactions

Consider a $0.1M$ solution of two solutes $A$ and $B$. $A$ behaves as a non-electrolyte while $80\%$ of $B$ dimerises. Which of the following statement is correct regarding these solutions?

  1. The b.pt of $A$ will be less than $B$
  2. The osomotic pressure of $B$ will be more than thatof $A$
  3. The freezing point of solution $A$ will be less than that of $B$

  4. Boiling points of both solutions will be same.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Correct option (b) The osmotic pressure of B will be more than that of A 

 Explanation: The concentration of electrolyte, which ionises in water shall be more although 80% of it dimerises 

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Water is boiled under a pressure of 1.0atm. When an electric current of 0.50A from a 12V supply is passed for 300 second through a resistance in thermal contact with it, it is found that 0.789g of water is vapourized. The molar internal energy change at boiling point (373.15K) is

  1. $\displaystyle =37.9kJ{ mol }^{ -1 }$
  2. $\displaystyle =27.5kJ{ mol }^{ -1 }$
  3. $\displaystyle =47.5kJ{ mol }^{ -1 }$
  4. $\displaystyle =17.5kJ{ mol }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vaporization occurs at constant pressure therefore the enthalpy change is equal to the work done by the heater:

$\displaystyle \Delta { H }^{ \prime  }=0.50\times 12\times 300\quad (\Delta H=i\times V\times t)$
$\displaystyle =1800J$
$\displaystyle =+1.8kJ$

$\displaystyle \therefore $Molar enthalpy of vaporization, $\displaystyle \Delta H=\frac { \Delta { H }^{ \prime  } }{ mole\ of\ { H } _{ 2 }{ O } } $
$\displaystyle =\frac { 1.8 }{ \left( \frac { 0.789 }{ 18 }  \right)  } $

$\displaystyle =41.06\ kJ{ mol }^{ -1 }$

Also, $\displaystyle \Delta H=\Delta U+P\Delta V$

$\displaystyle =\Delta U+\Delta { n } _{ g }RT$

$\displaystyle =\Delta U+RT$ $\displaystyle ((\because { H } _{ 2 }O\left( l \right) \rightleftharpoons { H } _{ 2 }O\left( g \right) ,\Delta { n } _{ g }=1)$

$\displaystyle \therefore \Delta U=$ molar internal energy change

$\displaystyle\Delta U= \Delta H-RT$

$\displaystyle =41.06-(8.314\times { 10 }^{ -3 }\times 373.15)$
$\displaystyle =37.96\ kJ{ mol }^{ -1 }$

Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

The saturated vapour pressure of water at $100^\circ C$ is

  1. 750 mm of Hg

  2. 760 mm of Hg

  3. 76 mm of Hg

  4. 7.6 cm of Hg
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vapour pressure or equilibrium vapour pressure or saturated vapour pressure is defined as the pressure exerted by a vapour in thermodynamic equilibrium with condensed phase at given temperature. At $100^oC$, the vapour pressure is $1atm$; i.e. 760mm of Hg i.e. 76cm of Hg as $1mm=10^{-1}cm$.

Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

Boiling point of water at normal atmospheric pressure is ______.

  1. $0^{\circ}C$
  2. $100^{\circ}C$
  3. $110^{\circ}C$
  4. $-4^{\circ}C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the environmental pressure surrounding the liquid. The boiling point of liquids varies with and depends upon the surrounding environmental pressure. 

The normal boiling of a liquid is the special case in which the vapor pressure of the liquid equals the defined atmospheric pressure at sea level i.e., at 1 atmosphere (atm) is $100^0C$.

Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

boiling point of petrol

  1. ${ 10 }^{ o }-{ 25 }^{ o }$
  2. $ { 120 }^{ o }-{ 180 }^{ o }C$
  3. $ { 30 }^{ o }-{ 120 }^{ o\quad }C$
  4. ${ 100 }^{ o }-{ 270 }^{ 0 }C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Petrol (gasoline) is a mixture of hydrocarbons with boiling points typically ranging from 30 to 200 degrees Celsius. Option C provides the most accurate range among the choices.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

For an ideal binary liquid solution with $P^{\circ} _{A} > P^{\circ} _{B}$, which relation between $X _{A}$ (mole fraction of A in liquid phase) and $Y _{A}$(mole fraction of $A$ in vapour phase) is correct?

  1. $Y _{A} < Y _{B}$
  2. $X _{A} > X _{B}$
  3. $\dfrac{Y _{A}}{Y _{B}} > \dfrac{X _{A}}{X _{B}}$
  4. $\dfrac{Y _{A}}{Y _{B}} < \dfrac{X _{A}}{X _{B}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an ideal binary liquid solution with $P _\overset {o}{A}>P _\overset {o}{B}$.

We know that, from Henry's law
$P _\overset {o}{A}\propto X _A$
So, $X _A > X _B$.
If mole fraction of $A$ in liquid phase is more then mole fraction of $A$ in vapour phase is less so, $Y _A < Y _B$.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

What is the molar mass of a non-ionizing solid if 10 g of this solid, dissolved in 100 g of water, formed a solution that froze at $-1.21^o C$?

  1. 0.65 g

  2. 65 g

  3. 130 g

  4. 154 g

  5. 265 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the freezing point depression formula: deltaTf = Kf * m. With Kf = 1.86, deltaTf = 1.21, and molality m = (10/M) / 0.1, solving for M gives approximately 154 g/mol.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

The vapour pressure of two pure liquids A and B are 200 and 400 torr respectively at 300K . A liquid solution (ideal) of A and B for which the mole fraction of A is 0.40 is contained in a cylinder. The composition of components A and B in vapour phase after equilibrium is reached between vapour & liquid phase, respectively is 

  1. $X _A = 0.62 ; X _B = 0.38$
  2. $X _A = 0.50 ; X _B = 0.50$
  3. $X _A = 0.25 ; X _B = 0.75$
  4. $X _A = 0.30 ; X _B = 0.70$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Raoult's Law: P_total = XA*PA + XB*PB. Then, the mole fraction in the vapor phase is YA = (XA*PA) / P_total. Given PA=200, PB=400, XA=0.4, XB=0.6, P_total = 0.4*200 + 0.6*400 = 80 + 240 = 320. YA = 80/320 = 0.25, YB = 0.75.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and in the vapour phases is 

  1. same

  2. different

  3. may or may not be same depending upon volatile nature of the two components

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For Solution $A+B$,

$X' _{A}=$ mole fraction of $A$ in vapour  phase

$X' _{A}=\dfrac{X _{A}.P^{0} _{A}}{X _{A}.P^{0}A}$ = can be equal to $X _{A}$ or not which depend on $P^{0} _{A}$ & $P^{0} _{B}$.
Multiple choice chemistry is matter pure different types of solutions various mixtures introduction to solutions

If air is taken as a binary solution, the solvent is:

  1. N$ _{ 2 }$
  2. O$ _{ 2 }$
  3. CO$ _{ 2 }$

  4. Ar
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) option is correct.

Explanation:
Air is mainly composed of oxygen and nitrogen gas. Around 78% is nitrogen gas and 21% is oxygen. Since the nitrogen content is more than oxygen so nitrogen will be the solvent and oxygen will be the solute if air is considered as binary solution.

Multiple choice common laboratory equipments common laboratory apparatus and equipments laboratory equipments know about some common gases chemistry

A solution of glucose received from some research laboratory has been marked mole fraction x and molality (m) at $1{ 0 }^{ \circ  }C$. When you will calculate its molality and mole fraction in your laboratory at $24^{ \circ  }C$ you will find:

  1. mole fraction (x) and molality (m)

  2. mole fraction (2x) and molality (2m)

  3. mole fraction (x/2) and molality (m/2)
  4. mole fraction (x) and (m$ _{ - }^{ + } $dm) molality
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molality and mole fractions both are independent of temperature . Therefore they will remain same after change of temperature.

Multiple choice chemistry separation of substances classification of mixtures mixtures: examples and properties types of solutions

An aqueous solution containing $1g$ of urea boils at $100.25^ {o}C$. The aqueous solution containing $3g$ of glucose in the same volume will boil at-

  1. $100.75^ {o}C$
  2. $100.5^ {o}C$
  3. $100^ {o}C$
  4. $100.25^ {o}C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The boiling point elevation depends on the molality of the solution. Urea (molar mass 60) and glucose (molar mass 180) have different molar masses. 1g urea is 1/60 moles, while 3g glucose is 3/180 = 1/60 moles. Since the number of moles is the same in the same volume, the molality is identical, resulting in the same boiling point elevation.

Multiple choice separation of components of mixtures methods of separation elements, compounds and mixtures chemistry

The vapour pressure of pure $A$ is $10$ torr and at the same temperature when $1\ g$ of $B$ is dissolved in $20\ g$ of $A$, its vapour pressure is reduced to $9.0$ torr. If the molecular mass of $A$ is $200\ amu$, then the molecular mass of $B$ is:

  1. $100\ amu$
  2. $90\ amu$
  3. $75\ amu$
  4. $120\ amu$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

So $0.1=\dfrac{moles \,\, of \,\, B}{moles of A+B}$

$0.1=\dfrac{1/m^2}{\dfrac{1}{m _2}+\dfrac{20}{200}}$
$0.1=\dfrac{1/m _2}{\dfrac{1}{m _2}+0.1}$
$\dfrac{0.1}{m _2}+0.01=\dfrac{1}{m _2}-\dfrac{0.1}{m _2}$
$0.01=\dfrac{0.9}{m _2}$
$m _2=\dfrac{0.9}{0.01}=90amu$