# How many grams of oxalic acid dihydrate (COOH)2 x 2H2O, is needed to prepare 525 mL of a .570 M solution?

Jul 10, 2017

#### Answer:

Approx. $38 \cdot g \ldots \ldots$

#### Explanation:

We require a MOLAR quantity of .............$525 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.570 \cdot m o l \cdot {L}^{-} 1 = 0.299 \cdot m o l$ with respect to $\text{oxalic acid dihydrate}$.

$\text{Oxalic acid dihydrate}$, $\text{HO"_2"CCO"_2"H"*"2H"_2"O}$, has a formula mass of $126.07 \cdot g \cdot m o {l}^{-} 1$.....

And thus we take the product....

126.07*g*mol^-1xx0.299*mol-=??*g.