# In order to make a 1*mol*L^-1 solution of NaNO_2 in water, how many grams of sodium nitrite are necessary?

Jan 21, 2016

Concentration $=$ $\left(\text{moles")/("volume of solution}\right)$. Here you need a $1$ $m o l \cdot {L}^{-} 1$ solution.

#### Explanation:

We require, $\frac{5 \cdot m m o l}{5 \cdot m L}$ $=$ $\frac{5 \times {10}^{- 3} \cdot m o l}{5 \times {10}^{-} 3 \cdot L}$ $=$ $1$ $m o l$ ${L}^{-} 1$

Note that the $m$ prefix simply means $\times {10}^{-} 3$. So $m m o l \cdot m {L}^{-} 1$ is precisely $1$ $m o l \cdot {L}^{-} 1$.

So to make this volume we simply take a $5$ $m L$ volume of $1$ $m o l \cdot {L}^{-} 1$ sodium nitrite.

It would be best to make a stock solution. Accurately weigh $70.00$ $g$ of sodium nitrite, and quantitatively transfer this to a $1$ $L$ volumetric flask. Fill up to the mark with distilled water, and make sure all the solid is dissolved.