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Question: Find 1.Mass of \[CaC{O_3}\] in its \[75\] milli-equivalents \[(meq.)\]. 2.Normality of the sol...

Find
1.Mass of CaCO3CaC{O_3} in its 7575 milli-equivalents (meq.)(meq.).
2.Normality of the solution in which 1.20×1024OH1.20 \times {10^{24}}\,O{H^ - } are present in 800mL800mL of the solution.

Explanation

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)(i) Milli-equivalents: It is the number of millimoles of the given compound multiplied by its normality.
Molar mass of CaCO3=100gCaC{O_3} = 100g
Calcium carbonate dissociates as follows:
CaCO3Ca2++CO32CaC{O_3} \to C{a^{2 + }} + C{O_3}^{2 - }
Therefore, n-factor =2 = 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=100250gCaC{O_3} = \dfrac{{100}}{2} \Rightarrow 50g
The equivalent mass we calculated is for one equivalent of CaCO3CaC{O_3}, so mass for 11 milli equivalents of CaCO3CaC{O_3} will be 50×103g50 \times {10^{ - 3}}g.
The mass for 7575 milli equivalents of CaCO3=75×50×103CaC{O_3} = 75 \times 50 \times {10^{ - 3}}
m=3.75g\Rightarrow m = 3.75g
Hence, the mass for 7575 milli equivalents of CaCO3=3.75gCaC{O_3} = 3.75g
(ii)(ii)In the question, the number of OHO{H^ - } ions are given, with the help of which we need to find the number of moles of OHO{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 OHO{H^ - } ions =1.2×10246.022×10231.99eq. = \dfrac{{1.2 \times {{10}^{24}}}}{{6.022 \times {{10}^{23}}}} \Rightarrow 1.99\,eq.
Volume of the solution given =800mL = 800mL
As 11 mL=103 = {10^{ - 3}}L
So, volume of solution in litres =800×1030.8L = 800 \times {10^{ - 3}} \Rightarrow 0.8L
Normality of the solution can be calculated as follows:
N=grameq.Volume(L)N = \dfrac{{gram\,eq.}}{{Volume(L)}}
N=1.990.82.48N\Rightarrow N = \dfrac{{1.99}}{{0.8}} \Rightarrow 2.48N
Hence, the normality of the solution is 2.48N2.48N

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.