How many moles of sodium are there in 10mls of 30% sodium chloride?

30% of 10mls is 3g NaCL. There is 3000mg of NaCl and would that be divided by the molecular weight of NaCl or of just Na?

I divided 3000mg of NaCl by 58.44 mg NaCl/mole and got 51.44 moles NaCl Since there is one mole of Na for every mole of NaCl there would be 51.44 moles of Na.

My teacher divided 3000mg (he didn't express what he had 3000mg of) by 22.99 mg Na/mole to get 130.5 moles of Na. I have found other similar problems on the internet and feel I am correct. My teachers only response was to read the question. If I am wrong please help me understand why.

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You are on the right track and with a few assumptions I'd say your calculation is correct. The one of your teacher is not.
Let's back up a moment and understand what is going on.

First of all we should make sure, that we are asking the right question. As you have given it, it seems incomplete. It should probably rather be:

What amount of substance of sodium (ions) can be found in 10 mL of a 30% aqueous solution of sodium chloride?

We have been given a mass fraction, which we need to convert to the amount of substance concentration first. The general formula for this is $$w_i = \frac{\rho_i}{\rho} = \frac{c_i M_i}{\rho},$$ where $\rho$ is the density of the solution, $\rho_i$ is the mass concentration of component $i$, $M_i$ is the molar mass of component $i$, and $c_i$ is the amount of substance concentration of $i$.

In your calculation you made the assumption, that the density of a 30% sodium chloride solution is $\rho = 1~\mathrm{g/mL}$, which is not correct. You can look up a couple of values for brine solutions on wikipedia.
For the purpose of this example, let's assume $\rho(\ce{NaCl~30\%}) \approx 1.2~\mathrm{g/mL}$.

The molar mass of sodium chloride is $M(\ce{NaCl}) = 58.4~\mathrm{g/mol}$. Be careful with the units, as you used the incorrect ones.

The amount concentration is given through the formula $$c_i = \frac{n_i}{V},$$ where $n_i$ is the amount of substance (not number of moles!) and $V$ is the volume of the solution.

You correctly identified that there is a one-to-one correlation between sodium ions and sodium chloride as expressed by $$\ce{NaCl (s) ->C[H2O] Na+ (aq) + Cl- (aq)}.$$ Therefore we can write $$n(\ce{Na+}) = n(\ce{NaCl}).$$

Putting everything together we arrive at the following formula: \begin{align} n(\ce{Na+}) &= \frac{w_{\ce{NaCl}}\cdot \rho(\ce{NaCl~30\%})}{M(\ce{NaCl})} \cdot V\\ &= \frac{0.3\cdot 1.2~\mathrm{g/mL}}{58.4~\mathrm{g/mol}}\cdot 10~\mathrm{mL}\\ &= 0.061~\mathrm{mol} \end{align}

Basically you thought process was correct under the assumption of the inaccurate density. And it was off by $10^{-3}$ because you (and your teacher) used the wrong units.

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  • $\begingroup$ I feel tempted to bounty this as an exemplary homework answer ;) $\endgroup$ – Jan Jun 24 '16 at 23:43
  • $\begingroup$ @Jan Thanks, but try to resist, I don't need the rep ;) $\endgroup$ – Martin - マーチン Jun 25 '16 at 9:02

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