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# 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.

Last updated date: 27th Mar 2023
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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.

$(i)$ 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.