One mole of chlorine combines with a certain mass of a metal giving $111\,g$ of its chloride. The same mass of metal can displace $2\,g$ of hydrogen from an acid. The atomic mass of the metal is,
A. $40$
B. $20$
C. $80$
D. None of the above
Answer
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Hint:We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molecular weight.
The mathematically expressed as,
${\text{Mole = }}\dfrac{{{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{substance}}}}{{\,{\text{Molecular}}\,{\text{weight}}}}$.
Complete step by step answer:
We know that the molecular formula of metal chloride is $MC{l_2}$.
A balanced chemical equation tells the number of moles of reactants and the number of moles of product.
First, write the balanced chemical equation of the reaction.
${M^{2 + }} + C{l_2}\xrightarrow{{}}MC{l_2}$
Given,
The weight of the metal chloride is $111\,g$.
We know, the atomic mass of chlorine is $35.5\,g.$
Calculate the molecular weight of the metal by using the above formula.
${\text{Mole = }}\dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 2{{\left( {{M_{wt}}} \right)}_{Cl}}}}$
${\text{Mole = }}\dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 71\,g}}\,\,\,\xrightarrow{{}}1$
Thus, $71\,g$ of chloride gives $111\,g$ of metal chloride that replaces $2\,g$ of hydrogen.
Now, metal can also displace two hydrogen atoms to the mole needed to displace hydrogen and one mole of chlorine should be the same.
${\text{Mole}}\,{\text{of the metal = }}\dfrac{2}{2} = 1\,\, \to 2$
Equate the equation 1 and 2,
$1 = \dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 71\,g}}$
${\left( {\,{{\text{M}}_{wt}}} \right)_{metal}} = 111 - 71$
${\left( {\,{{\text{M}}_{wt}}} \right)_{metal}} = 40$
The atomic mass of the metal is $40.$
$\therefore $Option A is correct.
Note:
We know that potassium and calcium have atomic masses of forty. The charge of calcium is $ + 2$ while the charge of potassium is $ + 1$. If the metal is potassium it uses two moles to replace hydrogen atom it gives final weight as $80$ but the actual weight obtained is 40. Hence the metal is calcium.
The mathematically expressed as,
${\text{Mole = }}\dfrac{{{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{substance}}}}{{\,{\text{Molecular}}\,{\text{weight}}}}$.
Complete step by step answer:
We know that the molecular formula of metal chloride is $MC{l_2}$.
A balanced chemical equation tells the number of moles of reactants and the number of moles of product.
First, write the balanced chemical equation of the reaction.
${M^{2 + }} + C{l_2}\xrightarrow{{}}MC{l_2}$
Given,
The weight of the metal chloride is $111\,g$.
We know, the atomic mass of chlorine is $35.5\,g.$
Calculate the molecular weight of the metal by using the above formula.
${\text{Mole = }}\dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 2{{\left( {{M_{wt}}} \right)}_{Cl}}}}$
${\text{Mole = }}\dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 71\,g}}\,\,\,\xrightarrow{{}}1$
Thus, $71\,g$ of chloride gives $111\,g$ of metal chloride that replaces $2\,g$ of hydrogen.
Now, metal can also displace two hydrogen atoms to the mole needed to displace hydrogen and one mole of chlorine should be the same.
${\text{Mole}}\,{\text{of the metal = }}\dfrac{2}{2} = 1\,\, \to 2$
Equate the equation 1 and 2,
$1 = \dfrac{{111}}{{{{\left( {\,{{\text{M}}_{wt}}} \right)}_{metal}} + 71\,g}}$
${\left( {\,{{\text{M}}_{wt}}} \right)_{metal}} = 111 - 71$
${\left( {\,{{\text{M}}_{wt}}} \right)_{metal}} = 40$
The atomic mass of the metal is $40.$
$\therefore $Option A is correct.
Note:
We know that potassium and calcium have atomic masses of forty. The charge of calcium is $ + 2$ while the charge of potassium is $ + 1$. If the metal is potassium it uses two moles to replace hydrogen atom it gives final weight as $80$ but the actual weight obtained is 40. Hence the metal is calcium.
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