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

1 Answer
May 20, 2015

You have all the information you need. "M" = ("mol")/"L"

M_"K" = "39.098 g/mol"
M_"N" = "14.007 g/mol"
M_"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")

= "66.72732 g"

The scales in my university measure to pm200 mu"g" of uncertainty ("0.0002 g"), so I would go for color(blue)("66.7273 g").

But, if you only need "1.1 M" and not "1.10 M" or "1.100 M", you can go for "66.73 g" and still be quite close.


NOTE:
In reality, that is a LOT of "KNO"_3. I don't think it would even fit a common plastic weigh boat, and I've weighed out about "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).

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Expect the volume of the water to almost double. When I weighed out "32 g" of urea in "25 mL" of water, it became about "50 mL" due to displacement.