Water has a density of 1 g/mL

$\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{}\overline{)\mathbf{L}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{g}}{\overline{)\mathbf{mL}}}\mathbf{=}$**1600 g**

The newest US standard for arsenate in drinking water, mandated by the Safe Drinking Water Act, requires that by January, 2006, public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. Parts per billion is defined on a mass basis as

**ppb = g solute/ g solution x 10 ^{9}**

If this arsenic is present as arsenate, AsO_{4}^{3 -}, what mass of sodium arsenate would be present in a 1.60 L sample of drinking water that just meets the standard? Express your answer using two significant figures.

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