# What is the molarity of .50 g of Na dissolved in a 1.5 L solution?

May 29, 2016

In aqueous solution, approx. $0.015 \cdot m o l \cdot {L}^{-} 1$ with respect to sodium ion.

#### Explanation:

$N a + {H}_{2} O \rightarrow N a O H \left(a q\right) + \frac{1}{2} {H}_{2} \left(g\right) \uparrow$

$\text{Moles of sodium metal}$ $=$ $\frac{0.50 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$

$\text{Concentration}$ $=$ $\text{Amount in moles"/"Volume of solution (L)}$

$=$ $\frac{0.50 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1.5 \cdot L}$

$=$ ??mol*L^-1.

Under standard conditions ($\text{298K, 1 atm}$), what volume of gas will be evolved?

Note that I could have dissolved the sodium in mercury metal; in fact this is often done in synthetic laboratories. Now that would be quite a heavy solution.