Question

A piece of ${\text{Mg}}$ is dissolved in $40{\text{ mL}} $ $\dfrac{{\text{N}}}{{{\text{10}}}}{\text{ HCl}}$ completely. The excess of acid was neutralised by $15{\text{ mL}}$of $\dfrac{{\text{N}}}{{\text{5}}}{\text{ NaOH}}$. Find the weight of Mg.
a)${\text{0}}{\text{.24g}}$
b) ${\text{0}}{\text{.024g}}$
c) ${\text{0}}{\text{.012g}}$
d) ${\text{0}}{\text{.40g}}$

Answer 1

:Hint:In the above question, we have to find out the weight of Mg which completely dissolves in the solution. Since, it is given that excess acid is neutralised by NaOH it implies that remaining acid reacted with Mg. To find out the weight of Mg, we have found the number of moles of Mg present which can be obtained by writing a balanced chemical equation between Mg and HCl.

Complete step by step answer:
Since it is given that ${\text{40 mL}}$ of $\dfrac{{\text{N}}}{{{\text{10}}}}$ HCl is present. To find the total number of moles of HCl present, we have to first convert normality to molarity. For converting normality to molarity, we have to divide the molarity by the number of hydrogen ions present in HCl. Since only one ${{\text{H}}^{\text{ + }}}$ ion is present so:
normality of HCl = molarity of HCl
So, ${\text{0}}{\text{.1N}}$HCl = ${\text{0}}{\text{.1M}}$HCl
From the definition of molarity we know that:
${\text{Molarity(M) = }}\dfrac{{{\text{number of moles of solute (n)}}}}{{{\text{Volume of solution in liter (V)}}}}$
Since, V=${\text{40mL = }}\dfrac{{{\text{40}}}}{{{\text{1000}}}}{\text{L = 0}}{\text{.04L}}$
So, number of moles of HCl (n1)=${\text{M \times V = 0}}{\text{.1 }} \times {\text{ 0}}{\text{.04 = 0}}{\text{.004 moles}}$
Similarly, for converting normality to molarity in NaOH, we have to divide the molarity by the number of ${\text{O}}{{\text{H}}^{\text{ - }}}$ present in NaOH. Since only one ${\text{O}}{{\text{H}}^{\text{ - }}}$ ion is present so
normality of NaOH = molarity of NaOH
So, $\dfrac{{\text{N}}}{{\text{5}}}{\text{ = 0}}{\text{.2N}}$ NaOH = ${\text{0}}{\text{.2M}}$ NaOH
As volume of NaOH is \[{\text{15mL = }}\dfrac{{{\text{15}}}}{{{\text{1000}}}}{\text{L = 0}}{\text{.015L}}\]
So, number of moles of NaOH (n2) = ${{M \times V = 0}}{\text{.015 }} \times {\text{ 0}}{\text{.2 = 0}}{\text{.003 moles}}$
Writing a balanced equation of NaOH and HCl:
${\text{NaOH + HCl }} \to {\text{ NaCl + }}{{\text{H}}_{\text{2}}}{\text{O}}$
It indicates that 1 mole of NaOH reacts with 1 mole of HCl. So, ${\text{0}}{\text{.03 moles}}$ of NaOH reacts with ${\text{0}}{\text{.03 moles}}$ of HCl.
So remaining moles of HCl =${\text{0}}{\text{.004 - 0}}{\text{.003 = 0}}{\text{.001 moles}}$
Writing a balanced equation between Mg and HCl, we get:
${\text{Mg + 2HCl }} \to {\text{ MgC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}$
It indicates that:
 2 moles of HCl reacts with 1 mole of Mg.
$ \Rightarrow {\text{0}}{\text{.001 moles}}$ of HCl reacts with $\dfrac{{0.001}}{2}{\text{ moles = 0}}{\text{.0005 moles}}$ of Mg.
So, the number of moles of Mg is ${\text{0}}{\text{.0005 moles}}$.
As molar mass of Mg is 24g
Mass of Mg= number of moles of Mg $ \times $Molar mass of Mg =$0.0005 \times 24{\text{ = 0}}{\text{.0012g}}$
$\therefore $Mass of Mg is ${\text{0}}{\text{.012g}}$.

So, option c is correct.

Note:
In these types of questions, we have to observe which values are given in the question. In this question, as volume was given so we had converted to molarity but if mass of the solvent would have given then we had to convert it to molality.