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If At 25 Degrees C, the molar solubility of barium chromate in water is 1.1 x 10^-5 M. Here is how you would find the solubility in grams per liter.

First: You must know the molar mass of Barium Chromate To continue. Look at your Periodic Table.

[BariumChromate Molar mass=253.37 g/mol]

*Get this by adding together atomic number of each element in BaCrO4

*253.37= Ba+Cr+O(4), or 137.33+52+16(4)

*Take 1.1 x 10^-5 M and divide by molar mass of Barium Chromate

1.1 x 10^-5 mol/L x 253.37 =0.002787

Answer in grams per liter to three significant figures=2.787 x 10^-3 g/L

How many liters of water are required to dissolve 1 of barium chromate?

*take 1gram BaCrO4 / 2.787 x 10^-3 g/L --------> Grams Cancel

=2.787 x 10^-3 L

Answer in liters to three significant figures =359.71 Liters

What is the solubility of barium chromate in parts per million?

*parts per million = Grams of Solute/grams of solution X 10^6 (which is ppm)

2.787 x 10^-3g/L x 1L/1000g x 10^6 = 0.02779, or 2.78 x 10^-2ppm

Answer in parts per million to three significant figures=2.78ppm

**For dilute aqueous solutions, ppm is equivalent to units of mg/l

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