Question

The number of moles of $S{{O}_{2}}C{{l}_{2}}$in $13.5gm$is [CPMT$1994$]
A.$0.1$
B.$0.2$
C.$0.3$
D.$0.4$

Answer 1

Hint: Mole is used for the measure of the amount of any substance. The number of moles of a substance equals the ratio of its given mass to the mass of one mole of that substance. Here we have also given the mass of $S{{O}_{2}}C{{l}_{2}}$and now only we have to calculate the molar mass of $S{{O}_{2}}C{{l}_{2}}$i.e mass of one mole $S{{O}_{2}}C{{l}_{2}}$.

Formula Used:Number of moles of any component,$n=\dfrac{m}{M}$
$m=$Given the mass of the component
$M=$mass of one mole of that component.

Complete answer:One mole of any component equals Avogadro’s number of particles which is $6.023\times {{10}^{23}}$ number of particles. It is also used as the unit of concentration such as $mol/liter$.
Here we have a compound,$S{{O}_{2}}C{{l}_{2}}$, so the molar mass of this compound is calculated below,
${{M}_{S{{O}_{2}}C{{l}_{2}}}}=$(Atomic weight of sulfur)$+$($2\times $Atomic weight of oxygen)$+$($2\times $Atomic weight of chlorine)
$\therefore {{M}_{S{{O}_{2}}C{{l}_{2}}}}=32+(2\times 16)+(2\times 35.5)=135gm/mol$
Given the mass of $S{{O}_{2}}C{{l}_{2}}$,${{m}_{S{{O}_{2}}C{{l}_{2}}}}$=$13.5gm$
The number of moles of $S{{O}_{2}}C{{l}_{2}}$,${{n}_{S{{O}_{2}}C{{l}_{2}}}}=\dfrac{{{m}_{S{{O}_{2}}C{{l}_{2}}}}}{{{M}_{S{{O}_{2}}C{{l}_{2}}}}}$
or,${{n}_{S{{O}_{2}}C{{l}_{2}}}}=\dfrac{13.5gm}{135gm/mol}=0.1mol$
Therefore, the number of moles of $S{{O}_{2}}C{{l}_{2}}$ is $0.1mol$.

Thus, option (A) is correct.

Additional information: Mole concept is very important in chemistry, as it is the foundation of stoichiometry. It is useful in expressing the reactants and products in any chemical reaction. All the chemical reactions can be expressed in terms of moles. One mole contains Avogadro’s number of particles, and particles may be atoms, molecules, electrons, protons, etc.

Note: To calculate the number of moles we should take care of its formula and the atomic weight from the periodic table. This is because sometimes atomic weights are not provided in the question but if we remember at least the atomic weight of twenty elements from hydrogen to calcium then it will be very helpful.