Tag: chemistry

Questions Related to chemistry

Dyes are used for colouring various materials like cloth and plastic. William Perkins, an English Chemist, discovered dyes for colouring from:

chemistry rocks and minerals coal and its products study of coal coal

  1. asphalt

  2. coal tar

  3. baking ovens

  4. zinc

Show answer
Correct Option: B
Explanation:
English chemist William Perkin discovered mauveine dye accidently when working with coal tar. Realising the commercial potential of this new purple dye, Perkin patented his synthetic process.

The main constituent of pitch during destructive distillation of coal tar is:

chemistry rocks and minerals coal and its products study of coal coal

  1. xylene

  2. anthracene

  3. carbon

  4. naphthalene

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Correct Option: C
Explanation:

The main constituent of pitch during destructive distillation of coal tar is carbon in various other forms.

Hence option C is correct.

The addition of $HCl$ will not suppress the ionisation of:

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. Acetic acid

  2. Sulphuric acid

  3. ${H} _{2}S$

  4. Benzoic acid

Show answer
Correct Option: B
Explanation:

The addition of $HCl$ will not suppress the ionisation of $H _2SO _4$ because $H _2SO _4$ is stronger acid than $HCl$.

One litre of water contains ${ 10 }^{ -7 }$ mole of ${H}^{+}$ ions. Degree of ionisation of water is:

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. $1.8\times { 10 }^{ -7 }$

  2. $0.8\times { 10 }^{ -9 }$

  3. $5. 4\times { 10 }^{ -9 }$

  4. $5 . 4\times { 10 }^{ -7 }$

Show answer
Correct Option: A
Explanation:

$H _2O⇌H^{ + }   +       OH^{ - }\quad $

  $C$        $0$                  $0$
$C(1-\alpha )$        $C\alpha $    $C\alpha $

$C\alpha=10^{-7}$

[H2O] =$\dfrac{ 1000}{18}$= 55.55 M 

C= 55.5M

So $\alpha=18\times10^{-10}$

In percentage $\alpha=1.8\times10^{-7}$%

When ${ NH } _{ 4 }Cl$ is added to ${ NH } _{ 4 }OH$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. common ion effect

  2. hydrolysis

  3. oxidation

  4. reduction

Show answer
Correct Option: A
Explanation:

When $NH _4Cl$ is added to $NH _4OH$ solution, concentration of $NH _4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced. 

The weak acid, $HA$ has a ${K} _{a}$ of $1.00\times { 10 }^{ -5 }$. If $0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. $1$%

  2. $99.9$%

  3. $0.1$%

  4. $99$%

Show answer
Correct Option: A
Explanation:
0.1 mole of acid is dissolved in 1 litre of water means $[HA]=0.1M$
Let '$\alpha$' be degree of dissocition
$HA\rightleftharpoons { H }^{ + }+{ A }^{ - }$
 $0.1$
$0.1(1-\alpha)$   $0.1\alpha$     $0.1\alpha$
${ K } _{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right]  }{ \left[ HA \right]  } =\cfrac { { 0.1 }^{ 2 }{ \alpha  }^{ 2 } }{ 0.1(1-\alpha)  } $
Let $\alpha<<1$ so $1-\alpha=1$
$K _a=0.1\alpha^2=10^{-5}$
$\alpha=10^{-2}$
% of acid dissociated=$1$%

If concentration of two acids are some, their relative strengths can be compared by:

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. ${ \alpha } _{ 1 }/{ \alpha } _{ 2 }$

  2. $K _{ 1 }/K _{ 2 }$

  3. ${ \left[ { H }^{ + } \right] } _{ 1 }/{ \left[ { H }^{ + } \right] } _{ 2 }$

  4. $\sqrt { K _{ 1 }/K _{ 2 } } $

Show answer
Correct Option: A,C,D
Explanation:

Relative strength of two acids can be compared by their degree of dissociation.

$HA\rightleftharpoons H^++A^-$
$C$
$C-C\alpha$   $C\alpha$    $C\alpha$
If concentration of two acids are same so their relative strength can be compared by their $[H^{+}]$ concentration.
$K _a=C\alpha^2$
$\alpha=(K _a/C)^{0.5}$
If concentration of two acids are same so their relative strength can be compared by square root of their dissociation constants.

The solubility of $AgCl$ in $NaCl$ solution is less than that in pure water, because of the  ________.

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. solubility product of $AgCl$ is less than of $NaCl$

  2. common ion effect

  3. both $A$ and $B$

  4. none of these

Show answer
Correct Option: B
Explanation:

Since, $NaCl$ is soluble to a very significant extent, when $AgCl$ is added to $NaCl$ solution, the common ion $[Cl^-]$ increases in the solution. To have the solubility product or $K _{sp}$ of $AgCl$ constant, $[Ag^+]$ will decrease or $AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of $AgCl$ in $NaCl$ solution will be less than that in pure water.

100 mL of 20.8% $BaCl _2$ solution and 50 mL of 9.8% $H _2SO _4$ solution will form $BaSO _4$
$(Ba=137, Cl=35.5, S=32, H=1, O=16)$
$BaCl _2+H _2SO _4\rightarrow BaSO _4+2HCl$

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. 23.3 g

  2. 11.65 g

  3. 30.6 g

  4. None of these

Show answer
Correct Option: B
Explanation:

$100ml$ of $20.8$% $BaCl _2$ solution= $20.8g$  $BaCl _2$

$50ml$ of $9.8$% $H _2SO _4$ solution= $4.9g$  $H _2SO _4$
Reaction: $BaCl _2+H _2SO _4\longrightarrow BaSO _4\downarrow +2HCl$
           $208 g mol^{-1}$   $98 g mol^{-1}$      $233 g mol^{-1}$
$\therefore 98g$ $H _2SO _4$ reacts with $208g$ $BaCl _2$
$4.9g$ $H _2SO _4$ reacts with $\cfrac {208}{98}\times 4.9=10.4 g$ $BaCl _2$
$98g$ $H _2SO _4$ will produce $233g$ $BaSO _4$
$\therefore 4.9g$ $H _2SO _4$ will produce= $\cfrac {233}{98}\times 4.9=11.65g$ $BaSO _4$

The addition of NaCl to AgCl decreases the solubility of AgCl because ________.

chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

  1. Solubility product decreases

  2. Solubility product remains constant.

  3. solution becomes unsaturated

  4. solution becomes super saturated.

Show answer
Correct Option: D
Explanation:

NaCl is highly soluble and when it is added to AgCl it decreases the solubility of AgCl because of common ion $Cl^-$ and solution become super saturated.