What mass of sodium iodide is required to prepare a $250.0mL$ volume of $1.50mol*L^-1$ concentration with respect to the salt?

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Answer 1 · 2017-04-08T17:17:05

Approx. $50*g$ ..............

By definition, $"concentration"="moles of solute"/"volume of solution"$

Here $"concentration"$ $=$ $1.5*mol*L^-1$ , and $"volume of solution"=250.0*mL$ or $250.0xx10^-3L$ .

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

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

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

$??*g$

What mass of sodium iodide is required to prepare a 250.0*mL250.0*mL volume of 1.50*mol*L^-11.50*mol*L^-1 concentration with respect to the salt?