Question

A mass of $0.355\;g$ of the compound ${{\text{M}}_2}C{O_3}.x{H_2}{\text{O}}$ (molar mass of ${\text{M}}\;:\;\;23\;gmo{l^{ - 1}}$) is dissolved in $100mL$ water and titrated against $0.05\;{\text{M}}\;HCl$ using methyl orange indicator. If the volume of $HCl$ consumed is $100mL$ the value of ${\text{x}}$ is:
A. $2$
B. $5$
C. $7$
D. $10$

Answer 1

Hint: Titration is the process of slow addition of one solution of known concentration to the other solution of known concentration until the solution becomes neutral. The solution which is added to another solution is known as titrant. The neutralization of solution is indicated by the change in colour of the solution. In this question we will see the reaction involved and calculate the molar mass of a given compound by calculating the molarity of each solution.

Complete answer:
Let’s write the reaction involved in the above process of titration
${{\text{M}}_2}C{O_3}\;\; + \;\;2HCl\; \to \;2MCl\;\; + \;{{\text{H}}_2}{\text{O}}\;\; + \;C{O_2}\;\;$
let’s write the values given to us
Mass of given compound, ${\text{w}}\; = \;0.355\;{\text{g}}\;\;$
Volume of Solvent(water), ${{\text{V}}_{\text{w}}}\; = \;100\;mL\;$
Volume of Hydrochloric acid, ${{\text{V}}_{HCl}}\; = \;100\;mL\;$
Molarity of Hydrochloric acid, ${{\text{M}}_{HCl}}\; = \;0.05\;mol{L^{ - 1}}$
Molar mass of Element, ${\text{m}}\; = \;23\;gmo{l^{ - 1}}$
Calculating number of moles of Hydrochloric acid from the formula
${\text{M}} = \dfrac{{\text{n}}}{{\text{V}}}$
Substituting values, we get
${\text{n}}\; = \;\dfrac{{\;0.05\; \times \;100}}{{1000}}$
${\text{n}}\; = \;0.005\;mol$
From the reaction we can see that two mol of Hydrochloric acid is reacting with one mol of compound so number of moles of compound in the reaction
${{\text{n}}_{\text{c}}}\; = \dfrac{{\;0.005\;}}{2}$
${{\text{n}}_{\text{c}}}\; = \;0.0025\;mol$
Now calculating the molecular weight of the compound by the formula of number of mole, the mass of x number of water molecules is taken as $18{\text{x}}$
${{\text{n}}_{\text{c}}}\; = \;\dfrac{{0.355}}{{\left( {23 \times 2 + 12 + 16 \times 3} \right) + 18{\text{x}}}}$
${{\text{n}}_{\text{c}}}\; = \;\dfrac{{0.355}}{{106 + 18{\text{x}}}}$
substituting the value of number of moles of compound
$0.0025\; = \;\dfrac{{0.355}}{{106 + 18{\text{x}}}}$
$.0025\; = \;\dfrac{{0.355}}{{106 + 18{\text{x}}}}$
we get
$18{\text{x}} = 36$
${\text{x}} = 2$
so, the value of x is $2$ and two molecules of water are there in the compound.

So, the correct answer is “Option A”.

Note:
Indicators are the substances whose solution changes colour due to change in the pH; they are generally weak acids or weak bases. The indicator used in the titration in this question is Methyl orange. It shows red colour in acidic medium and yellow colour in basic medium.