# How much NaNO_3 is needed to prepare 225mL of a 1.55 M solution of NaNO_3?

May 29, 2016

$\text{Approx. } 30 \cdot g$

#### Explanation:

$\text{Concentration}$ $=$

$\text{Mass of stuff (g)"/"Molar mass of stuff (mol)"xx1/"Volume of solution (L)}$

The calculation gives me units of $m o l \cdot {L}^{-} 1$ as required.

And the product,

$\text{Concentration"xx"Volume}$ $=$ $\text{Number of moles}$

$225 \times {10}^{-} 3 \cdot L \times 1.55 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.349 \cdot m o l$

$0.349 \cdot m o l \times 84.99 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g.

The calculation is consistent dimensionally.