Question

In 1L of pure water, ${\text{44}}{\text{.4g}}$of calcium chloride is dissolved. What is the number of ions present in 1mL of the resultant solution?
A.${\text{7}}{{.23 \times 1}}{{\text{0}}^{{\text{23}}}}$
B.${\text{7}}{{.23 \times 1}}{{\text{0}}^{{\text{20}}}}$
C.${\text{4}}{{.82 \times 1}}{{\text{0}}^{{\text{20}}}}$
D.${\text{4}}{{.82 \times 1}}{{\text{0}}^{{\text{23}}}}$

Answer 1

Hint:In the above question, we are asked to find out the number of ions present in 1mL of the resultant solution when ${\text{44}}{\text{.4g}}$ of calcium chloride is dissolved. First we have to find the number of molecules present in the solution using relation between moles and weight of the substance. Then, we can find out the number of ions present after seeing the number of ions present in a single molecule.

Formula Used-
${\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}$
where n = number of moles of the substance
            m =given mass of the substance
            M = molar mass of the substance

Complete step-by-step answer:In the question, ${\text{44}}{\text{.4g}}$of calcium chloride is used. So, first we have to find out the number of moles of calcium chloride present in the mass. For that, first we should find the molar mass of calcium chloride.
Formula of calcium chloride: ${\text{CaC}}{{\text{l}}_{\text{2}}}$
Molar mass of ${\text{CaC}}{{\text{l}}_{\text{2}}}$= ${1 \times }$atomic mass of calcium + ${2 \times }$atomic mass of chlorine
So, M =$20{{ + 2}} \times {\text{ 35}}{\text{.5 = 40 + 71 = 111g}}$
Given mass, m = ${\text{44}}{\text{.4g}}$
So, number of moles of ${\text{CaC}}{{\text{l}}_{\text{2}}}$=$\frac{{\text{m}}}{{\text{M}}}{\text{ = }}\frac{{{\text{44}}{\text{.4}}}}{{{\text{111}}}}{\text{ }} \approx {\text{ 0}}{\text{.4 moles}}$
So, in 1L solution, ${\text{0}}{\text{.4 moles}}$of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is present.
1000 mL solution contains ${\text{0}}{\text{.4 moles}}$ of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is present.
1 mL solution contains ${\text{0}}{\text{.0004 moles}}$ of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is present.
1 mole of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ contains ${\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}$
${\text{0}}{\text{.0004 moles}}$ of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ contains ${\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{ }} \times {\text{ 0}}{\text{.0004 molecules}}$
${\text{0}}{\text{.0004 moles}}$ of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ contains ${{24 \times 1}}{{\text{0}}^{19}}{\text{ molecules}}$
1 molecule of ${\text{CaC}}{{\text{l}}_{\text{2}}}$contains 2 ions ( ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$and ${\text{C}}{{\text{l}}^{\text{ - }}}$)
So, ${{24 \times 1}}{{\text{0}}^{19}}{\text{ molecules}}$of ${\text{CaC}}{{\text{l}}_{\text{2}}}$contains ${{2 \times 24 \times 1}}{{\text{0}}^{{\text{19}}}}{\text{ ions = 4}}{{.8 \times 1}}{{\text{0}}^{{\text{20}}}}{\text{ ions }}$
$\therefore $${\text{4}}{{.8 \times 1}}{{\text{0}}^{{\text{20}}}}{\text{ ions }}$are present in 1mL of resultant solution.

So, the correct option is option C.

Note:Avogadro’s number is kind of generalized number, hence, we can write:
${\text{1 mole}}$=$6.022 \times {10^{23}}$ atoms, or molecules, or protons, or electrons etc.
In short, it is the number of particles in a mole.
Hence, in order to solve these types of questions, we have to first decide which quantity (electron, proton, atom) is equivalent to Avogadro’s number.