What is the mass of calcium iodide required for a $2 \cdot L$ $2L$ volume of calcium iodide whose concentration is $2 \cdot m o l \cdot L^{- 1}$ $2mol*L^-1$ ?

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What is the mass of calcium iodide required for a $2 \cdot L$ $2L$ volume of calcium iodide whose concentration is $2 \cdot m o l \cdot L^{- 1}$ $2mol*L^-1$ ?

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Answer 1 · 2017-02-06T18:02:47

If you mean $calcium iodide$ $"calcium iodide"$ then you need a mass of $587.78 \cdot g$ $587.78*g$ .

Explanation:

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

And thus $moles of solute = concentration \times volume of solution$ $"moles of solute"="concentration"xx"volume of solution"$

And from the boundary conditions of the problem:

$moles of solute = concentration \times volume$ $"moles of solute"="concentration"xx"volume"$

$moles of solute = 2.0 \cdot m o l \cdot L^{- 1} \times 1 \cdot L = 2.0 \cdot m o l$ $"moles of solute"=2.0*mol*L^-1xx1*L=2.0*mol$ .

And $2 moles of calcium iodide$ $"2 moles of calcium iodide"$ represents a mass of $2 \cdot mol \times 293.89 \cdot g \cdot m o l^{- 1} = ? ? g .$ $2*"mol"xx293.89* g*mol^-1=??g.$

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What is the mass of calcium iodide required for a $2 \cdot L$ $2L$ volume of calcium iodide whose concentration is $2 \cdot m o l \cdot L^{- 1}$ $2mol*L^-1$ ?

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