Question

Find
1.Mass of \[CaC{O_3}\] in its \[75\] milli-equivalents \[(meq.)\].
2.Normality of the solution in which \[1.20 \times {10^{24}}\,O{H^ - }\] are present in \[800mL\] of the solution.

Answer 1

Hint: 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 \[CaC{O_3} = 100g\]
Calcium carbonate dissociates as follows:
\[CaC{O_3} \to C{a^{2 + }} + C{O_3}^{2 - }\]
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 \[CaC{O_3} = \dfrac{{100}}{2} \Rightarrow 50g\]
The equivalent mass we calculated is for one equivalent of \[CaC{O_3}\], so mass for \[1\] milli equivalents of \[CaC{O_3}\] will be \[50 \times {10^{ - 3}}g\].
The mass for \[75\] milli equivalents of \[CaC{O_3} = 75 \times 50 \times {10^{ - 3}}\]
\[ \Rightarrow m = 3.75g\]
Hence, the mass for \[75\] milli equivalents of \[CaC{O_3} = 3.75g\]
\[(ii)\]In the question, the number of \[O{H^ - }\] ions are given, with the help of which we need to find the number of moles of \[O{H^ - }\]. 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 \[O{H^ - }\] ions \[ = \dfrac{{1.2 \times {{10}^{24}}}}{{6.022 \times {{10}^{23}}}} \Rightarrow 1.99\,eq.\]
Volume of the solution given \[ = 800mL\]
As \[1\] mL\[ = {10^{ - 3}}\]L
So, volume of solution in litres \[ = 800 \times {10^{ - 3}} \Rightarrow 0.8L\]
Normality of the solution can be calculated as follows:
\[N = \dfrac{{gram\,eq.}}{{Volume(L)}}\]
\[ \Rightarrow N = \dfrac{{1.99}}{{0.8}} \Rightarrow 2.48N\]
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.