[sundrops]

After dissolving 1.00g of NaOH (40.g/mol) pellets in 5.0mL of water, what is the approximate molarity of the resulting solution?  


kay the way I went about solving this is:

1.00g NaOH / 40.0g/mol NaOH = 0.025 mol NaOH
0.025 mol NaOH / 0.005 L H20 = 5mol/L sol'n

but that seems too simple - is there a trick to this problem?
can I get your guys input?

[Borek]

Depends.

You are asked for approximate concentration, thus your result of 5M should do.

But there is a small trick. If you dissolve 1 g of NaOH in 5mL of water final volume will be not 5mL but slightly more. Thus real concentration will be below 5M - about 4.94M (CASC rulez). Thats 1% change, so for all practical purposes it doesn't matter.

[sundrops]

so I should take under consideration the extra 1ml of NaOH?

so my equation would look like:

0.0025mol Naoh / o.oo6L = 4.17 mol/L of sol'n?

thats a big difference btw 5M and 4.17 M.

[sundrops]

nevermind - ignore above post.

i see what you're saying borek - thank you!