Calculate the volume of a 0.068 M solution of zinc nitrate that would
provide 2.9 g of Zn(NO3)2.

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Clarification of Question by lilgreeny-ga on 06 Mar 2006 15:27 PST

I know that you have to do this first Calculate the molar mass of
Zn(NO3)2 to convert from grams to moles, which would be 2.9g x 1
mole/189.41g of Zn(NO3)2...right? ....but I do not know where to go
from there..   Please Help!!!

O.k., here is a logical solution that I came up with:

1st, convert the given mass of Zn(NO3)2 to moles
     2.9g *(1 mole/ 189.4g) = 0.0153 moles Zn(NO3)2

2nd, use this and you are given the concentration should be 0.068M (moles/ L)
     0.0153 moles*(1 L / 0.068 moles) = 0.225L = 225mL

225mL IS THE ANSWER!!!

This is logical because if you start off with 0.0153 moles, you need
less than one liter to get the resulting molarity.

Here is my answer:
number of moles= 2.9/189.41=0.0153 mol
Vol= number of moles/ conc= 0.0153/0.068= 0.225 l = 225 ml
Hence, I agree with the other answer given.