How many ions result from the dissolution of a $11.1 \cdot g$ $11.1*g$ mass of calcium chloride in water?

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How many ions result from the dissolution of a $11.1 \cdot g$ $11.1*g$ mass of calcium chloride in water?

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Answer 1 · 2017-03-27T14:24:30

Approx. $1.81 \times 10^{23}$ $1.81xx10^23$ ions

Explanation:

$Moles of calcium chloride$ $"Moles of calcium chloride"$ $=$ $=$ $\frac{11.1 \cdot g}{110.98 \cdot g \cdot m o l^{- 1}} = 0.100 \cdot m o l$ $(11.1*g)/(110.98*g*mol^-1)=0.100*mol$

And thus in solution:

$C a C l_{2} (s) + excess H_{2} O \to C a^{2 +} + 2 C l^{-}$ $CaCl_2(s) + "excess "H_2OrarrCa^(2+) + 2Cl^(-)$ , there are $0.100 \cdot m o l$ $0.100*mol$ $C a^{2 +}$ $Ca^(2+)$ ions, and $0.200 \cdot m o l$ $0.200*mol$ $C l^{-}$ $Cl^(-)$ ions, i.e. $0.300 \cdot m o l$ $0.300*mol$ of ions...........

And $0.300 \cdot m o l \times N_{A} = 0.300 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $0.300*molxxN_A=0.300*molxx6.022xx10^23*mol^-1$

$1.81 \times 10^{23}$ $1.81xx10^23$ ions..........

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How many ions result from the dissolution of a $11.1 \cdot g$ $11.1*g$ mass of calcium chloride in water?

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Explanation:

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How many ions result from the dissolution of a 11.1⋅g11.1*g mass of calcium chloride in water?

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How many ions result from the dissolution of a $11.1 \cdot g$ $11.1*g$ mass of calcium chloride in water?