Question
Question: Find 1.Mass of \[CaC{O_3}\] in its \[75\] milli-equivalents \[(meq.)\]. 2.Normality of the sol...
Find
1.Mass of CaCO3 in its 75 milli-equivalents (meq.).
2.Normality of the solution in which 1.20×1024OH− are present in 800mL of the solution.
Solution
Normality: It is a measure of concentration of a compound in terms of its equivalent mass. It is defined as the number of gram equivalents of solute or grams equivalent of solute per litre solution.
Complete answer:
(i) Milli-equivalents: It is the number of millimoles of the given compound multiplied by its normality.
Molar mass of CaCO3=100g
Calcium carbonate dissociates as follows:
CaCO3→Ca2++CO32−
Therefore, n-factor =2
As we know that the equivalent mass of any compound is the ratio of its molar mass to n-factor.
So, equivalent mass of CaCO3=2100⇒50g
The equivalent mass we calculated is for one equivalent of CaCO3, so mass for 1 milli equivalents of CaCO3 will be 50×10−3g.
The mass for 75 milli equivalents of CaCO3=75×50×10−3
⇒m=3.75g
Hence, the mass for 75 milli equivalents of CaCO3=3.75g
(ii)In the question, the number of OH− ions are given, with the help of which we need to find the number of moles of OH−. We know that the number of molecules or ions is the product of the number of moles and Avogadro’s constant.
So, number of moles (equivalents) of OH− ions =6.022×10231.2×1024⇒1.99eq.
Volume of the solution given =800mL
As 1 mL=10−3L
So, volume of solution in litres =800×10−3⇒0.8L
Normality of the solution can be calculated as follows:
N=Volume(L)grameq.
⇒N=0.81.99⇒2.48N
Hence, the normality of the solution is 2.48N
Note:
Equivalent mass: It is the mass of the compound which will exactly react to displace a fixed amount of another substance. N-factor is the maximum valency or charge that an element can carry. Equivalent mass is the ratio of molar mass and its n-factor.