Question

Suppose 7.17g of barium acetate is dissolved in 350mL of a 79.0mM aqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Round your answer to 1 significant digit.

Answer 1

Answer:

0.0802 M.

Explanation:

Hello!

In this case, since the barium acetate has the following formula:

$Ba(CH_3COO)_2$

So its molar mass is 255.43 g/mol, in order to compute the molarity of the barium cation, we first need the moles in 7.17 g:

$n_{Ba(CH_3COO)_2} = 7.17g*\frac{1molBa(CH_3COO)_2}{255.43g} =0.0281molBa(CH_3COO)_2$

Now, since one mole of barium acetate contains one mole of barium cations, we infer there are 0.0281 moles of barium cations. Moreover, since the molarity is computed by dividing the moles of those ones by the volume of the solution in liters, 350 mL (0.350 L) as it does not change, it turns out:

$M=\frac{0.0281molBa^{2+}}{0.350L}\\\\M=0.0802M$

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