What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?

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Answer 1 · 2017-04-12T04:34:39

1.49 M

Number of moles of sodium nitrate:
#n = m/(MW) = 11.8/85 = 0.13882 mol#

#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#

Answer 2 · 2017-04-12T05:20:15

#"Molarity"=1.48*mol*L^-1...................#

#"Molarity"="Moles of solute"/"Volume of solution"#.

For this problem.......

#"Molarity"=((11.8*cancelg)/(84.99*cancelg*mol^-1))/(93.9*cancel(mL)xx10^-3*L*cancel(mL^-1))#

#=??*mol*L^-1#...........