Question

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

Answer 1

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.

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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).

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.