Equilibrium Test

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Equilibrium Test - 36

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Question 1
1 / -0

The equivalent conductance of $$M/32$$ solution of monobasic acid is $$8.0$$ mho $$cm^2$$ and at infinite dilution is $$400$$ mho $$cm^2$$. The dissociation constant of this acid is:

A
$$1.25\times 10^{-5}$$

B
$$1.25\times 10^{-6}$$

C
$$6.25\times 10^{-4}$$

D
$$1.25\times 10^{-4}$$

Solution

Degree of dissociation,
$$\alpha=\frac{\lambda ^{c}}{\lambda ^{infinite}}$$

$$\alpha=\frac{8}{400} = 2\times 10^{^{-2}}$$

$$Ka = \frac{C\alpha ^{2}}{1-\alpha}$$

= $$C\alpha ^{2}$$ ( 1>>$$\alpha ^{2}$$)

= $$ 1.25\times 10^{-5}$$
Question 2
1 / -0

The total number of different kind of buffers obtained during the titration of $$H_3PO_4$$ with $$NaOH$$ are:

A
$$3$$

B
$$1$$

C
$$2$$

D
$$0$$

Solution

There will be 3 buffer solution they are
$$P{ O }_{ 4 }^{ 3- }$$- with $$NaOH$$
$$HP{ O }_{ 4 }^{ 2- }$$ with $$NaOH$$
$$HP{ O }_{ 4 }^{ - }$$ with $$NaOH$$
Question 3
1 / -0

At what temperature liquid water will be in equilibrium with water vapour?
$$\Delta H_{vap} = 40.73 \ kJ \ mol^{-1}$$, $$\Delta S_{vap} = 0.109 \ kJ \ K^{-1} \ mol^{-1}$$,

A
282.4 K

B
373.6 K

C
100 K

D
400 K

Solution

The given reaction is :-
$${ H }_{ 2 }O\left( l \right) \rightleftharpoons { H }_{ 2 }O\left( g \right) $$
Now, at equilibrium, $$\triangle G=0$$
Now, from fibb's free energy change equation
$$\triangle G=\triangle H-T\triangle S$$
$$\Rightarrow \quad \triangle H=T\triangle S$$
$$\Rightarrow \quad T=\dfrac { \triangle H }{ \triangle S } =\dfrac { 40.73 }{ 0.1079 } =373.6K$$
$$\left( \because \triangle Hrap=40.73KJ/mol,\quad \triangle Srap=0.109KJ{ K }^{ -1 }{ mol }^{ -1 } \right) $$
Question 4
1 / -0

What will be the value of pH of $$0.01 mol\ dm^{-3} CH_3COOH (K_a = 1.74 \times 10^{-5})$$?

A
$$3.4$$

B
$$3.6$$

C
$$3.9$$

D
$$3.0$$

Solution

$$\underset{C(1-\alpha)}{CH_3COOH}+H_2O\longrightarrow \underset{C\alpha}{H_3O}+\underset{C\alpha}{CH_3COO}$$
$$K_a=C\alpha^2\Rightarrow \alpha=\sqrt{\dfrac{Ka}{C}}$$
$$[H_3O^+]=\sqrt[C]{\dfrac{Ka}{c}}=\sqrt{Ka.C}$$
$$=\sqrt{0.01\times 1.74\times 10^{-5}}$$
$$\sqrt{1.74\times 10^{-7}}=4,17\times 10^{-4}$$
$$PH_2-\log(4.17\times 10^{-4})=3.38\simeq 3.4$$
Question 5
1 / -0

An acid $$HA$$ ionizes as, $$HA\rightleftharpoons H^++A^-$$. The pH of $$1.0$$M solution is $$5$$. Its dissociation constant would be:

A
$$1\times 10^{-10}$$

B
$$5$$

C
$$5\times 10^{-8}$$

D
$$1\times 10^{-5}$$

Solution

$$HA\rightleftharpoons H^+ +A^-$$
Initial $$\quad 1m\quad 0\quad 0$$
equilibrium $$1-x\quad x\quad x$$
given $$\left\{pH=5\right\}$$
$$5=-\log [H^+]$$
$$[H^+]=10^{-5}$$
So $$\left\{x=10^{-5}\right\}$$
$$Ka=\dfrac {[H^+] [A^-]}{[HA]}$$
$$Ka=\dfrac {(10^{-5} (10^{-5}))}{1-10^{-5}}$$
$$Ka=1\times 10^{-10} \left\{(1-10^{-5}\approx 1)\right\}$$
Question 6
1 / -0

At 373 K, steam and water are in equilibrium and $$\Delta H$$ = 40.98 kJ $$mol^{-1}$$.What will be $$\Delta S$$ fro conversion of water into steam?
$$H_2O_{(l)} \rightarrow H_2O_{(g)}$$

A
109.8 J $$K^{-1}mol^{-1}$$

B
31 J $$K^{-1}mol^{-1}$$

C
21.98 J $$K^{-1}mol^{-1}$$

D
326 J $$K^{-1}mol^{-1}$$

Solution

The given reaction is :-
$${ H }_{ 2 }O\left( l \right) \longrightarrow { H }_{ 2 }O\left( g \right) $$
So, we have
$$\triangle Svap=\dfrac { \triangle Hvap }{ Tb } $$
$$=\dfrac { 40.98\times 1000 }{ 373 } =109.8{ JK }^{ -1 }{ mol }^{ -1 }$$
Question 7
1 / -0

A buffer solution with $$ pH =9 $$ is to be prepared by mixing $$NH_4CI $$ and $$ NH_4OH $$. Calculate the number of moles of $$NH_4CI $$ that should be added to one litre pf $$1.0 M $$ $$ NH_4OH $$ . $$[K_b = 1.8 \times 10^{-5}]$$

A
$$3.4 $$

B
$$2.6 $$

C
$$1.5 $$

D
$$1.8 $$

Solution

Given $$pH=9, pOH=14-9=5$$
$$pOH=pK_b+\log \cfrac{[salt]}{[base]}$$
$$5=-\log(1.8\times 10^{-5})+\log\cfrac{[salt]}{1.0}=4.745+\log[salt]$$
$$\Longrightarrow [salt]=1.79\simeq 1.8$$
$$1.8$$ No. of moles of $$NH_4CI$$ to be added to $$1L$$ of $$1.0M$$ $$NH_4OH.$$
Question 8
1 / -0

Consider the reaction equilibrium:$$2SO_2+O_2\rightleftharpoons 2SO_3$$ : $$\Delta{H}^0$$= -195 kJ on the basis of Le Chateliers principle, the condition favourable for the forward reaction is?

A
Increasing temperature as well as pressure

B
Lowering the temperature and increasing the pressure

C
Any value of temperature and pressure

D
Lowering of temperature as well as pressure

Solution

$$ \begin{array}{l} 2SO_{2}+O_{2} \rightleftharpoons \quad 2 \mathrm{SO}_{3} \quad \Delta \mathrm{H}^{\circ}=-195 \mathrm{~K} \mathrm{~J} \\ \text { It is exothermic reaction } \\ \text { so lowering temperature favours forward } \\ \text { reaction according to lechatliers principle } \\ \end{array} $$
$$ \begin{array}{l} \Delta n=2-(2+1)=-1 \\ \text { no. of moles decreases then increase in } \\ \text { Pressure favours forward reaction. } \\ \text { so (B) is correct } \end{array} $$
Question 9
1 / -0

$$2HI \rightleftharpoons H_2 + I_2$$
The equilibrium constant of the above reaction is $$6.4$$ at $$300$$ K. If $$0.25$$ mole each of $$H_2$$ and $$I_2$$ are added to the system, the equilibrium constant will be:

A
1.6

B
3.2

C
0.8

D
6.4
Question 10
1 / -0

Strong electrolyte of the following is?

A
$$01M$$ $$HAc$$

B
$$0.1M$$ $$HCl$$

C
$$0.1M$$ $$KCl$$

D
$$0.1M$$ $$NaCl$$

Solution

B is correct
Because it dissociates completely.

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Equilibrium Test - 36

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Equilibrium Test - 36

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Question 1

$$1.25\times 10^{-5}$$

$$1.25\times 10^{-6}$$

$$6.25\times 10^{-4}$$

$$1.25\times 10^{-4}$$

Solution

Question 2

$$3$$

$$1$$

$$2$$

$$0$$

Solution

Question 3

282.4 K

373.6 K

100 K

400 K

Solution

Question 4

$$3.4$$

$$3.6$$

$$3.9$$

$$3.0$$

Solution

Question 5

$$1\times 10^{-10}$$

$$5$$

$$5\times 10^{-8}$$

$$1\times 10^{-5}$$

Solution

Question 6

109.8 J $$K^{-1}mol^{-1}$$

31 J $$K^{-1}mol^{-1}$$

21.98 J $$K^{-1}mol^{-1}$$

326 J $$K^{-1}mol^{-1}$$

Solution

Question 7

$$3.4 $$

$$2.6 $$

$$1.5 $$

$$1.8 $$

Solution

Question 8

Increasing temperature as well as pressure

Lowering the temperature and increasing the pressure

Any value of temperature and pressure

Lowering of temperature as well as pressure

Solution

Question 9

1.6

3.2

0.8

6.4

Question 10

$$01M$$ $$HAc$$

$$0.1M$$ $$HCl$$

$$0.1M$$ $$KCl$$

$$0.1M$$ $$NaCl$$

Solution

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