How much #NaNO_3# is needed to prepare 225mL of a 1.55 M solution of #NaNO_3#?

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Answer 1 · 2016-05-29T06:04:33

#"Approx. " 30* g#

#"Concentration"# #=#

#"Mass of stuff (g)"/"Molar mass of stuff (mol)"xx1/"Volume of solution (L)"#

The calculation gives me units of #mol*L^-1# as required.

And the product,

#"Concentration"xx"Volume"# #=# #"Number of moles"#

#225xx10^-3*Lxx1.55*mol*L^-1# #=# #0.349*mol#

#0.349*molxx84.99*g*mol^-1# #=# #??g#.

The calculation is consistent dimensionally.