Tag: chemistry

Questions Related to chemistry

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atoms. At $0^o C$ and $1\ atm$, its density is $1.25\ g$ per litre. The formula of the gas would be ______________.

  1. $CH _2$

  2. $C _2H _4$

  3. $C _2H _6$

  4. $C _4H _8$


Correct Option: B
Explanation:

$PV=nRT$


$P=\dfrac {dRT}{M}$

$M=\dfrac {1.25\times 0.082\times 273}{1}$

$=27.98$

Molar mass $\equiv 28\ gram/mole$

$\therefore \ $ as empirical formula $=CH _2$

$\therefore \ $ emprical mass $=12+2=14$

$n=\dfrac {28}{14}=2$

$\therefore \ $ Molecular formula $=(CH _2) _2$

$=C _2H _4$

Hence, the correct option is $\text{B}$

When burnt in air, $14.0\ g$ mixture of carbon and sulpher gives a mixture of $CO 2$ and $SO _2$ in the volume ratio of $2:1$, volume being measured at the same conditions of temperature and pressure. Moles of carbon in the mixture is _________.

  1. $0.25$

  2. $0.40$

  3. $0.5$

  4. $0.75$


Correct Option: C
Explanation:

Consider the mass of $C=x$


mass of $S=14-x$

moles of $C=\dfrac {x}{12};$ moles of $\dfrac {14-x}{32}$

$C+O _2\to CO _2;\quad S+O _2\to SO _2$

moles of $C=$ moles of $CO _2=\dfrac {x}{12}\quad $ moles of $S=$ moles of $SO _2=\dfrac {14-x}{32}$

Given,

$\dfrac {V _c}{V _s}=\dfrac {2}{1}$

$V\alpha $ no. of mole

$\dfrac {\dfrac {x}{12}}{\dfrac {14-x}{32}}=\dfrac {2}{1}$

$\dfrac {x}{12}\times \dfrac {32}{14-x}=\dfrac {2}{1}$

$\dfrac {4x}{42-3x}=1$

$4x=42-3x$

$7x=42$ 

$\boxed {x=6}$

$\therefore \ $ moles of $C$ in mixture $=\dfrac {w}{M}=\dfrac {6}{12}=0.5$

Answer is option $C$

If air is pumped slowly but continuously into a metallic cylinder of strong wall, what would happen to the air inside the cylinder?

  1. Temperature of air would increase

  2. Pressure of air would increase

  3. Pressure of air would decrease

  4. Temperature and pressure of air would increase


Correct Option: B
Explanation:

Due to continuous pumping number of moles will increase
We know,
$n\propto P$
$n\propto V$
$n\propto \dfrac{1}{T}$
$\therefore$ Pressure will increase.
Option B.

For $10$ min each, at $27^o$C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of $3$l capacity. The resulting pressure is $4.18$ bar and the mixture contains $0.4$ mole of nitrogen. The molar mass of the unknown gas is?

  1. $112$g $mol^{-1}$

  2. $242$g $mol^{-1}$

  3. $224$g $mol^{-1}$

  4. $422$g $mol^{-1}$


Correct Option: C

What is the molecular formula of arsenious chloride?

  1. $ AsCI _{3}$

  2. $ As _{2}CI _{6}$

  3. $ As _{2}CI _{5}$

  4. $ AsCI _{5}$


Correct Option: B

If two compounds have the same empirical formula but different molecular formula, they must have :

  1. different percentage composition

  2. different molecular weights

  3. same vapour density

  4. none of these


Correct Option: B
Explanation:

If two compounds have the same empirical formula but different molecular formula, they must have different molecular weights.


For example, $CH _2O$ and $C _6H _{12}O _6$ have the same empirical formula but different molecular formula, they have different molecular weights.

Option B is correct.

Calculate the number of molecules present in $0.5$ moles of magnesium oxide $\left( MgO \right) $. 

[Atomic weights :  $Mg=24, O=16$]

  1. $14.09\times { 10 }^{ 23 }$ molecules

  2. $3.0115\times { 10 }^{ 23 }$ molecules

  3. $30.12\times { 10 }^{ 23 }$ molecules

  4. $14.09\times { 10 }^{ -23 }$ molecules


Correct Option: B
Explanation:
1 mole of MgO contains Avogadro's number of molecules which is $ \displaystyle 6.023 \times 10^{23}$ molecules.

0.5 mole of MgO will contain $ \displaystyle 0.5 \times 6.023 \times 10^{23}=3.0115 \times 10^{23}$ molecules.

Common salt is chemically sodium chloride ($NaCl$).

  1. True

  2. False


Correct Option: A
Explanation:

The well-known table salt is chemically sodium chloride. Its chemical formula is $NaCl$.

The product of electrolysis of conc. solution of common salt is?

  1. $Na+Cl _{2}$

  2. $H _{2}+O _{2}$

  3. $NaOH+H _{2}+Cl _{2}$

  4. $Na+Cl _{2}+O _{2}$


Correct Option: C
Explanation:

The ionization of salt:

$NaCl\rightarrow Na^{+}+Cl^{-}$

The half-reaction at the cathode is:

$H _2O+2e^{−}\rightarrow H _2+2OH^{−}$

Here, the reduction is taking place.

The half-reaction at the anode is:

$2Cl^{-}\rightarrow Cl _2+1e^{−}$

Here, oxidation is taking place.

The overall reaction is as follows:

$NaCl+H _2O\rightarrow Na^{+}+OH^{−}+H _2+Cl _2$.

Hence, option $C$ is correct.

Which of the following is not hydroscopic-

  1. NaCl

  2. ${\text{MgC}}{{\text{l}} _{2\;\;}}$

  3. ${\text{CaC}}{{\text{l}} _{2\;\;}}$

  4. LiCl


Correct Option: D