Tag: masses of atoms and molecules

Questions Related to masses of atoms and molecules

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$=4\,Mg/mol$