Question

Suppose 2.11 g of barium nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of sodium chromate.


Suppose 2.11 g of barium nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of sodium chromate. 


Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. 

Round your answer to 2 significant digits. 

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Answer #1

Moles of barium nitrate = ( mass /molar mass)

= (2.11/261.32)

= 0.008.

Volume of solution = 50 mL = 5 × 10-3 L

Ba(NO3)2 (aq)  + Na2CrO4 (aq)  \to BaCrO4 (s) + 2 NaNO3 (aq)

So, moles of nitrate ion = 2 × moles of Barium nitrate

= 2× 0.008

= 0.016.

Molarity of nitrate ion

= ( 0.016/5×10-3)

= 3.2 M. ( 2 significant digit)

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