Tag: various mixtures

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Questions Related to various mixtures

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

Which of the following sequences lists the relative sizes of particles in a water mixture from smallest to largest?

  1. Solutions, suspensions, colloids

  2. Solutions, colloids, suspensions

  3. Colloids, solutions, suspensions

  4. Colloids, suspensions, solutions

  5. Suspensions, colloids, solutions
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The increasing order for the relative particle size is solutions ( 0.1 to 1 ) < colloids  (1 to 1000 ) < suspensions (more than 1000). The numbers in parenthesis refer to particle size in nm.
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

A solution is defined as a:

  1. homogeneous mixture of two or more substances

  2. hetrogeneous mixture of two or more substances

  3. homogeneous mixture of liquid and solid components only

  4. homogeneous mixture consisting of water as one of the components
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A solution is a homogeneous mixture of two or more substances. A solution may exist in any phase. For example, in a saline solution, salt is the solute dissolved in water as the solvent.

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

Arrange the following in the increasing order of their solubility in n-octane based on solute-solvent interaction: 

  1. $KCl < CH _3CN < CH _3OH < Cyclohexane$
  2. $KCl < Cyclohexane < CH _3OH < CH _3CN$
  3. $KCl < CH _3OH < CH _3 CN < Cyclohexane$
  4. $ KCl < Cyclohexane < CH _3CN < CH _3CN$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solubility in n-octane (a non-polar solvent) follows the 'like dissolves like' principle. KCl is ionic (least soluble), followed by polar molecules, with non-polar cyclohexane being most soluble.

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

an organic liquid (A), containing C,H and O with boiling point $78^{\circ}$C and possessing a rather pleasant odour or heating with concenterated $H _{2}SO _{4}$ gives a gaseous product (B) with the empirical formula $CH _{2}$. B decolourises bromine water as well as alkaline $KMnO _{4}$ solution and takes up one mole of $H _{2}$(per mole of B) in the presence of finely divided nickle at high temperature. A and B are:

  1. $C _{2}H _{5}OH$, $C _{2}H _{2}$
  2. $CH _{3}OH$, $C _{2}H _{2}$
  3. $C _{2}H _{5}OH$, $C _{2}H _{4}$
  4. $(CH _{3}) _{2}CHOH$, $C _{3}H _{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$C _2H _5OH$ or ethyl alcohol when treated with concentrated $H _2SO _4$ undergoes dehydration to form ethene($C _2H _4$). Ethane reacts with alkaline $KMnO _4$ undergoes oxidation to form ethylene glycol. Ethene  undergoes hydrogenation in presence of nickel heads of one mole of $H _2$ to form ethane.

The correct option is C

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

A sample of $CH _4$ of 0.08 g was subjected to combustion at $27^oC$ in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by $0.25^oC$. If heat capacity of calorimeter is 18 kJ, $\Delta H$ for combustion of $CH _4$ at $27^oC$ is:

  1. $- 900$ kJ/mole
  2. $- 905$ kJ/mole
  3. $- 895$ kJ/mole
  4. $- 890$ kJ/mole
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ CH } _{ 4 }\left( g \right) +{ 2O } _{ 2 }\left( g \right) \longrightarrow { CO } _{ 2 }\left( g \right) +2{ H } _{ 2 }O\left( l \right) $

$\Delta E=$ Heat of combustion
        $=$ Heat capacity $\times$ rise in $T$ $\times$ $\dfrac { Molar\ mass }{ Mass\ of\ compound } $
        $=-18\times 0.25\times \dfrac { 16 }{ 0.08 } $
        $=-900$ KJ/mole                                         $R=8.314\times { 10 }^{ -3 }KJ{ mol }^{ -1 }$
$\Delta H=\Delta E+\Delta nRT$                                      $\Delta n=1-3=-2$
$=-900+\left( -2 \right) \times 8.314\times { 10 }^{ -3 }\times 300$           $T=300K$
$=-900-4.9884=-904.98\simeq -905KJ/mol$

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

$CoCl _{ 3 }.3NH _{ 3 }$ does not form any precipitate with excess ${ AgNO } _{ 3 }$ solution, whereas 1 mole of $CoCl _{ 3 }.5NH _{ 3 }$ gives two moles of $AgCl$ with excess ${ AgNO } _{ 3 }$. The van't Hoff factor for both the compounds respectively are:

  1. 0 and 2

  2. 0 and 3

  3. 1 and 3

  4. 1 and 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To give ppt with ${ AgNO } _{ 3 }$ solution, a substance has to release ${ Cl }^{ - }$ ions.
$CoCl _{ 3 }.3NH _{ 3 }$ gives no ppt because it does not dissociate in solution.
$\therefore $ Number of particles in solution $=1$ i.e the molecule itself $CoCl _{ 3 }.5NH _{ 3 }$ gives 2 moles $AgCl$ ppt because it dissociate as follows
$CoCl _{ 3 }.5NH _{ 3 }\rightarrow \left[ Co\left( { NH } _{ 3 } \right) _{ 5 }Cl \right] ^{ 2+ }+2{ Cl }^{ - }$
It gives 3 ions on dissociation.
$\therefore $ its van't Hoff factor is 3.

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 $P^{0} _{A}$ & $P^{0} _{B}$.