# Question dc400

May 18, 2017

Dissolve 27.6 g of $\text{NaH"_2"PO"_4·"H"_2"O}$ in enough water to make 1 L of solution.

#### Explanation:

Let's say you want to prepare 1 L of 0.200 mol/L ${\text{NaH"_2"PO}}_{4}$ solution from solid $\text{NaH"_2"PO"_4·"H"_2"O}$.

That means you want to add 0.200 mol of the solid to enough water to make 1 L.

The molar mass of $\text{NaH"_2"PO"_4·"H"_2"O}$ is 137.99 g/mol.

So, you would measure out

0.200 color(red)(cancel(color(black)("mol NaH"_2"PO"_4·"H"_2"O" )))× ("137.99 g NaH"_2"PO"_4·"H"_2"O")/(1 color(red)(cancel(color(black)("mol NaH"_2"PO"_4·"H"_2"O")))) = "27.6 g NaH"_2"PO"_4·"H"_2"O"#.

You would transfer this carefully to a 1 L volumetric flask and use about 500 mL of distilled water to dissolve the solid.

Then you would add enough more distilled water to make the volume up to the 1 L mark.