
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
Answer
582.3k+ views
Hint: 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 ${\rm{Molarity}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Volume}}\;{\rm{of}}\,{\rm{solution\; in\; litre}}}}$.
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,
${\rm{Moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar\; mass}}}}$ …… (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, ${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}}$.
${\rm{Molar}}\;{\rm{mass}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}} = 2 \times 12\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 1\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 16\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
$ = 24\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 96\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
$ = 126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
Now, we know the molar mass and mass of oxalic acid. Molar mass is $126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$and mass is 12.6 g. So, we use equation (1) to calculate moles of oxalic acid in the solution.
${\rm{Moles}}\;{\rm{of}}\;{\rm{oxalic}}\;{\rm{acid}} = \dfrac{{{\rm{Mass}}\;{\rm{of\;oxalic\; acid}}}}{{{\rm{Molar}}\;{\rm{Mass}}\;{\rm{of\; oxalic \;acid}}}}$
$ = \dfrac{{12.6\;{\rm{g}}}}{{126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}$
$ = 0.1\;{\rm{mol}}$
Now we know the moles of solute, that is, oxalic acid in the solution. Volume of the solution is $100\;{\rm{ml}}$. But to calculate molarity, we need the volume in litre. So, we have to convert volume to litre.
${\rm{Volume}} = 100\;{\rm{ml}} \times \dfrac{{1\;{\rm{L}}}}{{1000\,{\rm{ml}}}}$
$ = 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.
${\rm{Molarity}} = \dfrac{{{\rm{Moles}}\;{\rm{of\; oxalic\; acid}}}}{{{\rm{Volume}}}}$
$ = \dfrac{{0.1\;{\rm{mol}}}}{{0.1\;{\rm{L}}}}$
$ = 1\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}$
$ = 1\;{\rm{M}}$
So, molarity of solution is $1\;{\rm{M}}$. 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.
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,
${\rm{Moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar\; mass}}}}$ …… (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, ${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}}$.
${\rm{Molar}}\;{\rm{mass}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}.2{{\rm{H}}_{\rm{2}}}{\rm{O}} = 2 \times 12\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 1\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 16\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
$ = 24\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 96\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
$ = 126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
Now, we know the molar mass and mass of oxalic acid. Molar mass is $126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$and mass is 12.6 g. So, we use equation (1) to calculate moles of oxalic acid in the solution.
${\rm{Moles}}\;{\rm{of}}\;{\rm{oxalic}}\;{\rm{acid}} = \dfrac{{{\rm{Mass}}\;{\rm{of\;oxalic\; acid}}}}{{{\rm{Molar}}\;{\rm{Mass}}\;{\rm{of\; oxalic \;acid}}}}$
$ = \dfrac{{12.6\;{\rm{g}}}}{{126\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}$
$ = 0.1\;{\rm{mol}}$
Now we know the moles of solute, that is, oxalic acid in the solution. Volume of the solution is $100\;{\rm{ml}}$. But to calculate molarity, we need the volume in litre. So, we have to convert volume to litre.
${\rm{Volume}} = 100\;{\rm{ml}} \times \dfrac{{1\;{\rm{L}}}}{{1000\,{\rm{ml}}}}$
$ = 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.
${\rm{Molarity}} = \dfrac{{{\rm{Moles}}\;{\rm{of\; oxalic\; acid}}}}{{{\rm{Volume}}}}$
$ = \dfrac{{0.1\;{\rm{mol}}}}{{0.1\;{\rm{L}}}}$
$ = 1\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}$
$ = 1\;{\rm{M}}$
So, molarity of solution is $1\;{\rm{M}}$. 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

