Please solve it

Dear Student,

Please find below the solution of your asked query:

$$\begin{gathered} \mathrm{The} \mathrm{relation} \mathrm{between} \mathrm{molarity} \mathrm{and} \mathrm{mole} \mathrm{fraction} \mathrm{is} \\ {\mathrm{X}}_{{\mathrm{CaCl}}_2} =\frac{{\mathrm{MM}}_{{\mathrm{H}}_2\mathrm{O}}}{\mathrm{M}({\mathrm{M}}_{{\mathrm{H}}_2\mathrm{O}}-{\mathrm{M}}_{{}_{{\mathrm{CaCl}}_2}})+\mathrm{d}} \\ \mathrm{Where} {\mathrm{X}}_{{}_{{\mathrm{CaCl}}_2}} = \mathrm{mole} \mathrm{fraction} \mathrm{of} {\mathrm{CaCl}}_2 \\ \mathrm{M}= \mathrm{Molarity} = \mathrm{To} \mathrm{be} \mathrm{calculated} \\ {\mathrm{M}}_{{\mathrm{H}}_2\mathrm{O}} = \mathrm{Molecular} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 18 \mathrm{g}/\mathrm{mol} \\ {\mathrm{M}}_{{\mathrm{CaCl}}_2} = \mathrm{Molecular} \mathrm{mass} \mathrm{of} {\mathrm{CaCl}}_2 = 111 \mathrm{g}/\mathrm{mol} \\ \mathrm{d} = \mathrm{Density} \mathrm{of} \mathrm{solution} = 1.025 \mathrm{g}/\mathrm{ml} \\ \mathrm{Now}, \mathrm{substituting} \mathrm{the} \mathrm{values} \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{mentioned} \mathrm{formula}: \\ 0.013=\frac{\mathrm{M}\times 18}{\mathrm{M}\left(18-111\right)+1.025} \end{gathered}$$


After solving it, we get,

M = 6.92 * 10^-4 Molar

Hope this information clears your doubts about the topic.

Keep asking!!

Regards