Question

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Suppose 0.193 g of copper(II) nitrate is dissolved in 350. mL of a 17.0 m M aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) nitrate is dissolved in it. Be sure your answer has the correct number of significant digits. Ом 01

Answer 1

Given, Molarity of sodium chromate = 17.0 mM volume of sodium chromate solution = 350.0 mL mass of copper (II) nitrate (Cu(NO3)2) = 0.193 g molar mass of Cu(NO3) = 187.56 g/mol calculate the moles of Cu(NO3), as follows: mass of Cu(NO) moles of Cu(NO3) = molar mass of Cu(NO) 0.193 g 187.56 g/mol = 0.00103 mol - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Write theionisation of Cu(NO3), as follows: Cu(NO3), -→ Cu2+ + 2NO3 moles of Cu2+ = moles of Cu(NO), = 0.00103 mol molarity is defined as the number of moles of solute present is 1 L volume of solution. moles of solute molarity, M = - Volume of solution Calculate the molarity of Cu2+ as follows: molarity of Cu2+ moles of Cu2+ Volume of solution 0.00103 mol 1000 mL 350.0 mL 1L = 0.00294 M = 2.94x10-3M Therefore, the required solution is 2.94 x10-2 M