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Question: Molarity of solution 12.6 grams of oxalic acid in 100 ml of the solution is: A. 0.75 M B. 0.5 M ...

Molarity of solution 12.6 grams of oxalic acid in 100 ml of the solution is:
A. 0.75 M
B. 0.5 M
C. 1.0 M
D. 2 M

Explanation

Solution

Molarity is one way to calculate concentration. It is the ratio of moles of solute to volume of solution in litre. The formula of molarity is Molarity=Moles  of  soluteVolume  of solution  in  litre{\rm{Molarity}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Volume}}\;{\rm{of}}\,{\rm{solution\; in\; litre}}}}.

Complete step by step answer:
In the question, it is given that the solution is an oxalic acid solution. So, solute is oxalic acid. To calculate molarity, we need the moles of solute. To calculate moles, we need mass of solute and molar mass of solute. The formula to calculate moles is,

Moles=MassMolar  mass{\rm{Moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar\; mass}}}} …… (1)

First we calculate the moles of oxalic acid. Mass of oxalic acid given is 12.6 g. Now, we need to calculate the molar mass of oxalic acid.
To calculate the molar mass of oxalic acid, we need chemical formula of oxalic acid, that is, C2H2O4.2H2O{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}}.

Molar  mass  C2H2O4.2H2O=2×12  g  mol−1+6×1  g  mol−1+6×16  g  mol−1{\rm{Molar}}\;{\rm{mass}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}} = 2 \times 12\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 1\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 16\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}
=24  g  mol−1+6  g  mol−1+96  g  mol−1= 24\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 96\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}
=126  g  mol−1= 126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}

Now, we know the molar mass and mass of oxalic acid. Molar mass is 126  g  mol−1126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}and mass is 12.6 g. So, we use equation (1) to calculate moles of oxalic acid in the solution.

Moles  of  oxalic  acid=Mass  of  oxalic  acidMolar  Mass  of  oxalic  acid{\rm{Moles}}\;{\rm{of}}\;{\rm{oxalic}}\;{\rm{acid}} = \dfrac{{{\rm{Mass}}\;{\rm{of\;oxalic\; acid}}}}{{{\rm{Molar}}\;{\rm{Mass}}\;{\rm{of\; oxalic \;acid}}}}
=12.6  g126  g  mol−1= \dfrac{{12.6\;{\rm{g}}}}{{126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}
=0.1  mol= 0.1\;{\rm{mol}}

Now we know the moles of solute, that is, oxalic acid in the solution. Volume of the solution is 100  ml100\;{\rm{ml}}. But to calculate molarity, we need the volume in litre. So, we have to convert volume to litre.

Volume=100  ml×1  L1000 ml{\rm{Volume}} = 100\;{\rm{ml}} \times \dfrac{{1\;{\rm{L}}}}{{1000\,{\rm{ml}}}}
=0.1 L= 0.1\,{\rm{L}}

So, the volume of the solution is 0.1 L. Now, we use the value of moles of oxalic acid and volume of solution to calculate molarity.

Molarity=Moles  of  oxalic  acidVolume{\rm{Molarity}} = \dfrac{{{\rm{Moles}}\;{\rm{of\; oxalic\; acid}}}}{{{\rm{Volume}}}}
=0.1  mol0.1  L= \dfrac{{0.1\;{\rm{mol}}}}{{0.1\;{\rm{L}}}}
=1  mol  L−1= 1\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}
=1  M= 1\;{\rm{M}}

So, molarity of solution is 1  M1\;{\rm{M}}. Hence, option C is correct.

Note: Students might confuse molarity and molality. Molarity is the moles of solute present in litres of solution and molality is the moles of solute present in per kilogram of solvent.