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.......

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

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

What is the molarity of a solution of NaNO_3NaNO_3 (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?