Tag: chemistry

Questions Related to chemistry

The number of gram-molecules of oxygen in $6.022\times 10^{24}$ molecules of $CO$ is:

  1. $10$ gm moles

  2. $5$ gm moles

  3. $1$ gm mole

  4. $0.5$ gm mole


Correct Option: B
Explanation:

Number of oxygen atoms = Number of $CO$ molecules$=6.022\times 10^{24}$

Number of oxygen molecule $=\dfrac12\times$  Number of oxygen atoms $=3.011\times 10^{24}$
Number of g-molecule of $O _2$ molecules $=\dfrac{3.011\times 10^{24}}{6.022\times 10^{23}}=5\ gm\ mole$

The molar mass of $CuSO _4.5H _2O$ is 249. Its equivalent mass in the reaction (a) and (b) would be:
(a) Reaction $CuSO _4 + KI \rightarrow$ product
(b) Electrolysis of $CuSO _4$ solution

  1. (a) 249 (b) 249

  2. (a) 124.5 (b) 124.5

  3. (a) 249 (b) 124.5

  4. (a) 124.5 (b) 249


Correct Option: B
Explanation:

In water, ${ CuSO } _{ 4 }\cdot 5{ H } _{ 2 }O$ dissociates into ${ Cu }^{ 2+ }$ & ${ SO } _{ 4 }^{ 2- }$. The ions are doubly charged. Hence half as much is needed to react with a compound that dissociates into singly charged species.

(a) Reaction of ${ CuSO } _{ 4 }+KI\longrightarrow $ products.
     Eq. wt of ${ CuSO } _{ 4 }=249/2=124.5$
(b) Electrolysis of ${ CuSO } _{ 4 }$
     At Cathode ${ \underset { +2 }{ Cu }  }^{ 2+ }+{ 2e }^{ - }\longrightarrow \underset { 0 }{ Cu } $
$\therefore$   Charge in oxidation state $=2=n-$factor
$\therefore$   Eq. wt $=\dfrac { 249 }{ 2 } =124.5$

10 ml of 0.1 M solution sodium hydroxide is completely neutralised by 25 ml of 3 gram of dibasic acid in one solution the molecular weight of acid is:

  1. 225 g

  2. 250 g

  3. 300 g

  4. 150 g


Correct Option: D
Explanation:
Given,
Concentration of $NaOH=0.1 N$
Volume$=10\,mL$
Mass of Dibasic acid$=3g$
Volume$=25\,mL$
Using,
$n _{1}M _{1}V _{1}=n _{2}M _{2}V _{2}$
$n _{2}=2$
$\Rightarrow 1\times 0.1\times 10=2\times25\times\cfrac{3}{M}$
$\Rightarrow M=\cfrac{2\times 25\times 3}{10\times 0.1}$
$\Rightarrow M=150\,g$

M g of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecular mass of the substance will be: 

  1. $M$

  2. $2M$

  3. $3M$

  4. $4M$


Correct Option: D
Explanation:
Given,
$Mg$ of substance occupy volume$=5.6\,litre $ at NTP.
At NTP,
1 mol occupy 22.4 litre of volume.
5.6 litre$=Mg$
22.4 litres$=4\,Mg$ of substance.
So, Molecular mass of gas$=4\,Mg/mol$

Gaseous $ N _{2}O _{4} $ dissociates into gaseous $ NO _{2} $ according to the reaction $ N _{2}O _{4} (g) \rightleftharpoons 2NO _{2}(g)$ at 300 K and 1 atm pressure, the degree of dissociation of $ N _{2}O _{4} $ is 0.2. If one mole of $ N _{2}O _{4} $ gas is contained in a vessel, then the density of the equilibrium mixture is : 

  1. 3.11 g/L

  2. 4.56 g/L

  3. 1.56 g/L

  4. 6.22 g/L


Correct Option: A
Explanation:

$N _2O _4\longrightarrow 2NO _2$


at $t=0$, moles of $N _2O _4=1$, moles of $NO _2=0$

at $t=equilibrium$, mole of $N _2O _4=1-a$, mole of $NO _2=2a$

$a$=degree of dissociation.

Molecular weight of mixture$=\cfrac {(1-a)\times\text{molar mass of }N _2O _4+2a\times \text{molar mass of }NO _2}{(1-a+2a)}$
                                                $=\cfrac {(1-0.2)(28+64)+2\times 0.2\times (14+32)}{1+0.2}$
$M=76.66$

$P=1 atm,T=300K,$

$d=PM/RT$

    $=\cfrac {1\times 76.66}{0.082 \times 300}=3.11\  gm/lit\ $ .

Attainment of equilibrium in a coloured gaseous reversible reaction is detected by the constancy of:

  1. colour

  2. density

  3. pressure

  4. all the above properties of the mixture


Correct Option: D
Explanation:

The intensity of color represents the concentration of either reactant or product. Thus when in a colored gaseous reversible reaction, the color has attained constant intensity, the concentrations of reactants and products have reached equilibrium values. In other words, an equilibrium is attained.


Hence, the correct option is A.

$4g \,H _2$ and $127g \,I _2$  are mixed and heated lit closed vessels until equilibrium is reached. If the equilibrium concentration of $HI$ is $0.05 \,M$ total number of moles present at equilibrium is:

  1. $3.25$

  2. $1.75$

  3. $2.25$

  4. $2.5$


Correct Option: D
Explanation:

Given $H _2=4g$ & $I _2=127 g$

$\Rightarrow H _2=2$ mole & $I _2=0.5$ mole
$[HI] _{eqm}=0.05M$   $\therefore$ Moles of $HI=0.05$ mole
            $H _2\quad +\quad I _2\quad \rightleftharpoons\quad  2HI$
              $2$              $0.5$                  $0$         Initial
  $2-\cfrac {0.05}{2}$     $0.5-\cfrac {0.05}{2}$       $0.05$       Eqm
$\therefore$ Total moles at eqm,
$=\left(2-\cfrac {0.05}{2}\right)+\left(0.5-\cfrac {0.05}{2}\right)+0.05$
$=2.5$

For the reaction ${ CO(g)+H } _{ 2 }O(g)\rightleftharpoons { CO } _{ 2 }(g)+{ H } _{ 2 }(g)$ at a given temperature the equilibrium amount of ${ CO } _{ 2 }(g)$ can be increased by:

  1. Adding a suitable catalyst

  2. Adding an inert gas

  3. Decreasing the volume of container

  4. Increasing the amount of $CO(g)$


Correct Option: D
Explanation:

$CO _(g)^+\ H _2(g)\rightleftharpoons CO _{2(g)}+H _{2(g)}$

$\Delta x=0$   $\therefore$ Adding inert gas & decreasing volume will have no effect. by increasing amount of CO, we shift reaction forward and to more $CO _2$.

Match the items in column - I with those in column - II

Column I Column II
1. Electric Fuse  Chemical Effect
2. Relay B Electric Discharge
3. CFL C Magnetic effect 
4. Button Cell D Heating Effect

Which of the following shows the correct matching ?

  1. 1 - C, 2 - B, 3 - A, 4 - D

  2. 1 - B, 2 - A, 3 - C, 4 - D

  3. 1 - D, 2 - C, 3 - B, 4 - A

  4. 1 - D, 2 - B, 3 - C, 4 - A


Correct Option: C
Explanation:

Electric fuse is used to protect electric appliances. When current larger than specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.


 In relay transmission electromagnetic waves are used. 

CFL works on electric discharge. In button cell electrolytes are the sources of ions.

In the button cell electrolytes are sources of ions.

Option C is correct.

Equilibrium can be achieved only in open vessel.
  1. True

  2. False


Correct Option: B
Explanation:

Equilibrium can be achieved in open vessel as well as in close vessel.