If 500 mL of CaCl_2 solution contains 3.01 t | vedclass

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(A) The dissociation of $CaCl_2$ is given by: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$. One mole of $CaCl_2$ produces $2$ moles of chloride ions $(Cl^-)$.

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(A) The dissociation of $CaCl_2$ is given by: $CaCl_2 ightarrow Ca^{2+} + 2Cl^-$. One mole of $CaCl_2$ produces $2$ moles of chloride ions $(Cl^-)$. Number of moles of $Cl^-$ ions $= \frac{3.01 	imes 10^{22}}{6.02 	imes 10^{23}} = 0.05 \ mol$. Since $2$ moles of $Cl^-$ come from $1$ mole of $CaCl_2$,the moles of $CaCl_2 = \frac{0.05}{2} = 0.025 \ mol$. Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.025 \ mol}{0.5 \ L} = 0.05 \ M$. Thus,the molarity of the solution is $0.05 \ M$.

If $500 \ mL$ of $CaCl_2$ solution contains $3.01 \times 10^{22}$ chloride ions,the molarity of the solution will be: (in $M$)

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