The number of Cl^{-} ions in 333 g of anhydr | vedclass

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Question

(A) The molar mass of $CaCl_2$ is $40 + 2 	imes 35.5 = 111 \ g/mol$. Number of moles of $CaCl_2 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{333 \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The molar mass of $CaCl_2$ is $40 + 2 	imes 35.5 = 111 \ g/mol$. Number of moles of $CaCl_2 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{333 \ g}{111 \ g/mol} = 3 \ mol$. Each molecule of $CaCl_2$ dissociates to give $2 \ Cl^{-}$ ions. Therefore,$3 \ mol$ of $CaCl_2$ contains $3 	imes 2 = 6 \ mol$ of $Cl^{-}$ ions. The number of $Cl^{-}$ ions is $6 \ N_A$.

The number of $Cl^{-}$ ions in $333 \ g$ of anhydrous $CaCl_2$ will be ................ $N_A$.

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