Question #22084

What is the molarity of a solution that contains 0.350 g of barium fluoride, BaF2, in 0.50 L of solution?

Expert's answer

Molarity or C = mol per L

So you need to find amount of BaF_{2}:

n = m/Mw ( Mw = 175)

n = 0.350/175 = 0.002 mol

So it is 0.002 mol per 0.50 L and concentration is two

much more becose it is some amount per one L

C = 0.004 M

So you need to find amount of BaF

n = m/Mw ( Mw = 175)

n = 0.350/175 = 0.002 mol

So it is 0.002 mol per 0.50 L and concentration is two

much more becose it is some amount per one L

C = 0.004 M

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