When $7.1gm$$N{{a}_{2}}S{{O}_{4}}$(molecular mass $142$) dissolves in $100ml$${{H}_{2}}O$, the molarity of the solution is [CBSE PMT $1991$; MP PET$1993,95$]
A.$2.0M$
B.$1.0M$
C.$0.5M$
D.$0.05M$

Hint: The molar concentration of any solution can be expressed in terms of molarity. Molarity is nothing but the number of moles present per liter of solution. In this question, the molecular mass of the solute has already been provided we just have to calculate the number of moles $N{{a}_{2}}S{{O}_{4}}$and put the given values in the formula of molarity given in below.

Formula Used:(i)Number of moles of a substance,$n=\dfrac{m}{M}$
Here $m=$mass of the substance
And $M=$the molecular mass in $gm/mol$
(ii)Molarity,$M=\dfrac{n}{V(litre)}$
Where $n$denotes the number of moles and $V$is the volume in $litre$.

Complete answer:Here the solute $N{{a}_{2}}S{{O}_{4}}$is dissolved in a solvent $100\,ml$ ${{H}_{2}}O$. The number of moles of a component equals the ratio of its given mass in a reaction to the mass of one mole of that component which is molecular mass.
Then the number of moles of $N{{a}_{2}}S{{O}_{4}}$, ${{n}_{N{{a}_{2}}S{{O}_{4}}}}=\dfrac{{{m}_{N{{a}_{2}}S{{O}_{4}}}}}{Molecular\,\,mass}$
Given the mass of $N{{a}_{2}}S{{O}_{4}}$,${{m}_{N{{a}_{2}}S{{O}_{4}}}}$$=7.1gm$
And molecular mass$=142\,gm/mol$
$\therefore {{n}_{N{{a}_{2}}S{{O}_{4}}}}=\dfrac{7.1gm}{142\,gm/mol}=0.05\,mol$
Molarity of the solution,$M=\dfrac{{{n}_{N{{a}_{2}}S{{O}_{4}}}}}{V}$ …….(i)
Given the volume of solution,$V=100\,ml$
As we know $1\,litre=1000\,ml$
So, the volume in $litre$$=0.1\,litre$
Putting the above values in equation (i),
$\therefore M=\dfrac{0.05\,mol}{0.1\,litre}=0.5\,mol/litre$
Therefore the molarity of the given solution is $0.5\,\,mol/litre$.

Thus, option (C) is correct.

Additional information: A solution is formed when one component dissolves into another. A solution is homogeneous when a solute is dissolved into a solvent. Therefore solute is defined by the substance that is being dissolved whether the solvent is a medium in which solute is dissolved. For example, in a solution of sugar and water, sugar is the solute and water is the solvent.

Note: To approach these types of problems one must have an idea about mole concepts such as molarity, normality, molality, etc, and also have to focus on unit conversion. In molarity, the solvent is taken in $litre$but in molality, the solvent is taken in $kg$.