In a $1 \cdot m L$ $1mL$ volume of solution that is $0.10 \cdot m o l \cdot L^{- 1}$ $0.10mol*L^-1$ with respect to $C a C l_{2} (a q)$ $CaCl_2(aq)$ , what is the NUMBER of $ions in solution?$ $"ions in solution?"$

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In a $1 \cdot m L$ $1mL$ volume of solution that is $0.10 \cdot m o l \cdot L^{- 1}$ $0.10mol*L^-1$ with respect to $C a C l_{2} (a q)$ $CaCl_2(aq)$ , what is the NUMBER of $ions in solution?$ $"ions in solution?"$

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Answer 1 · 2017-06-11T14:08:23

There are $3 \times 10^{- 4} \cdot m o l \times N_{A}$ $3xx10^-4*molxxN_A$ ions in solution.......

Explanation:

Now $concentration$ $"concentration"$ is given by the quotient:

$concentration = \frac{Moles of solute}{Volume of solution}$ $"concentration"="Moles of solute"/"Volume of solution"$

And thus to get the $moles of solute$ $"moles of solute"$ we takes the product

$Volume of solution \times concentration$ $"Volume of solution "xx" concentration"$ .......

And so here,

$moles of calcium chloride$ $"moles of calcium chloride"$ $=$ $=$

$1 \cdot m L \times 10^{- 3} L \cdot m L^{- 1} \times 0.10 \cdot m o l \cdot L^{- 1} = 10^{- 4} \cdot m o l$ $1*mLxx10^-3L*mL^-1xx0.10*mol*L^-1=10^-4*mol$

WITH RESPECT TO THE SALT $C a C l_{2}$ $CaCl_2$ .

Why did we emphasize $the salt$ $"the salt"$ ? Because EACH formula unit of $C a C l_{2}$ $CaCl_2$ contributes 3 ions to the solution, i.e.....

$C a C l_{2} (s) H_{2} O - - \to C a^{2 +} + 2 C l^{-}$ $CaCl_2(s)stackrel(H_2O)rarrCa^(2+) + 2Cl^-$

And so we have $3 \times 10^{- 4} \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1} = ? ?$ $3xx10^-4*molxx6.022xx10^23*mol^-1=??$ ions??

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In a $1 \cdot m L$ $1mL$ volume of solution that is $0.10 \cdot m o l \cdot L^{- 1}$ $0.10mol*L^-1$ with respect to $C a C l_{2} (a q)$ $CaCl_2(aq)$ , what is the NUMBER of $ions in solution?$ $"ions in solution?"$

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In a 1⋅mL1*mL volume of solution that is 0.10⋅mol⋅L−10.10*mol*L^-1 with respect to CaCl2(aq)CaCl_2(aq), what is the NUMBER of ions in solution?"ions in solution?"

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In a $1 \cdot m L$ $1mL$ volume of solution that is $0.10 \cdot m o l \cdot L^{- 1}$ $0.10mol*L^-1$ with respect to $C a C l_{2} (a q)$ $CaCl_2(aq)$ , what is the NUMBER of $ions in solution?$ $"ions in solution?"$