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

1 Answer
May 29, 2016

Answer:

#"Approx. " 30* g#

Explanation:

#"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.