# A laboratory procedure calls for making 600.0mL of a 1.1M KNO_3 solution. How much KNO_3 in grams is needed?

May 20, 2015

You have all the information you need. $\text{M" = ("mol")/"L}$

${M}_{\text{K" = "39.098 g/mol}}$
${M}_{\text{N" = "14.007 g/mol}}$
${M}_{\text{O" = "15.999 g/mol}}$

M_("KNO"_3) = 39.098 + 14.007 + 3*15.999 = "101.102 g/mol"

(1.1 cancel("mol"))/cancel("L") * (1cancel("L"))/(1000cancel("mL")) * 600.0 cancel("mL") * (101.102 "g")/cancel("mol")

$=$ $\text{66.72732 g}$

The scales in my university measure to $\pm 200$ $\mu \text{g}$ of uncertainty ("0.0002 g"), so I would go for $\textcolor{b l u e}{\text{66.7273 g}}$.

But, if you only need $\text{1.1 M}$ and not $\text{1.10 M}$ or $\text{1.100 M}$, you can go for $\text{66.73 g}$ and still be quite close.

NOTE:
In reality, that is a LOT of ${\text{KNO}}_{3}$. I don't think it would even fit a common plastic weigh boat, and I've weighed out about $\text{32 g}$ of solid urea before. And even then, it may take at least $5$ minutes of constant swishing around to mix (unless you have a sonicator). Expect the volume of the water to almost double. When I weighed out $\text{32 g}$ of urea in $\text{25 mL}$ of water, it became about $\text{50 mL}$ due to displacement.