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

Question

4351 views around the world

You can reuse this answer
Creative Commons License

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*mL# volume of #1.50*mol*L^-1# concentration with respect to the salt?