HELP ME PLEASE CHEM QUESTION

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ngernqjryy29

Science

Description

Suppose 2.25 g of sodium chloride is dissolved in 50mL of a .20M aqueous solution of ammonium sulfate.

Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium chloride is dissolved in it.

Round your answer to 2 significant digits.

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Explanation & Answer

initial mole of NaCl = 0.20M x 50x10^-3 L = 0.01 mol

molar mass of NaCl = 58.44 g/mol

additional mole = 2.25g / 58.44 g/mol = 0.0385 mol

final mole = 0.01 mol + 0.0385 mol = 0.0485 mol

since mole of NaCl = mole of Na+

The final molarity of sodium cation in the solution is then calculated by:

0.0485 mol / 50x10^-3 L = 0.97 M

Answer: 0.97 M


Anonymous
Excellent resource! Really helped me get the gist of things.

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