Question
Question: The mass of oxalic acid \(({{H}_{2}}{{C}_{2}}{{O}_{4}}\cdot 2{{H}_{2}}O)\) required to prepare 50ml ...
The mass of oxalic acid (H2C2O4⋅2H2O) required to prepare 50ml of a 0.2N solution is:
[A] 4.5g
[B] 6.3g
[C] 0.63g
[D] 0.45g
Solution
Hint: To find the required mass, at first we need to calculate the molar mass of oxalic acid. The required mass can be found out by multiplying the molar mass by the quantity and the molarity of the solution.
Complete answer:
The formula of oxalic acid is (H2C2O4⋅2H2O)
Atomic weight of Hydrogen=1
Atomic weight of Carbon=12
Atomic weight of Oxygen=16
Therefore, the molar mass of oxalic acid= (1×2)+(12×2)+(16×4)+(2×1)+(2×16)=126
50ml = 50÷100=0.05L
Normality is the gram equivalents of solute present per litre of solution and molarity is the grams of solute per litre of solution. To find the amount of NaOH added in grams, we have to convert normality into molarity.
Here, we can use the formula, Molarity= Normality ÷ acidity
Acidity can be written as the number of hydronium ions released per mole in a solution.
∴ We have to convert 0.2N to molarity.
0.2N means 0.2 gram equivalent is present in 1L of the solution.
Since, oxalic acid is a dibasic acid because it furnishes two H+ ions per molecule in water.
∴0.2N=0.2÷2molar=0.1molar
The required mass of oxalic acid= 0.1×126×0.05g=0.63g
Therefore, 0.63g of oxalic acid is needed to prepare a 50ml 0.2N solution of oxalic acid.
Therefore, the correct answer is option C, 0.63g.
ADDITIONAL INFORMATION:
This sort of calculation is very common during practical experiments where a certain molar solution is needed to carry out the experiment.
Note: It is important to remember here to convert normality into molarity and millilitres into litres as we will get wrong results if we do not keep a check on the unit.
It is also important to remember the atomic number of the elements in order to find out the molar mass. Generally, atomic weight is the double of atomic number. E.g. Atomic number of carbon= 6, Atomic weight of carbon= 12. However, it is not true for every element but it is almost equal to the double of atomic number.