What mass of sodium iodide is required to prepare a $250.0 \cdot m L$ $250.0mL$ volume of $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ concentration with respect to the salt?

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What mass of sodium iodide is required to prepare a $250.0 \cdot m L$ $250.0mL$ volume of $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ concentration with respect to the salt?

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Answer 1 · 2017-04-08T17:17:05

Approx. $50 \cdot g$ $50*g$ ..............

Explanation:

By definition, $concentration = \frac{moles of solute}{volume of solution}$ $"concentration"="moles of solute"/"volume of solution"$

Here $concentration$ $"concentration"$ $=$ $=$ $1.5 \cdot m o l \cdot L^{- 1}$ $1.5*mol*L^-1$ , and $volume of solution = 250.0 \cdot m L$ $"volume of solution"=250.0*mL$ or $250.0 \times 10^{- 3} L$ $250.0xx10^-3L$ .

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

$=1.5*mol*cancel(L^-1)xx0.250*cancelL=0.375*mol$ of $"NaI"$ .

And mass of $"NaI"$ required $=$ $0.375*molxx149.89*g*mol^-1$

$??*g$

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What mass of sodium iodide is required to prepare a $250.0 \cdot m L$ $250.0mL$ volume of $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ concentration with respect to the salt?

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