HELP ME PLEASE CHEM QUESTION

Science
Tutor: None Selected Time limit: 1 Day

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.
Oct 3rd, 2015

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

Sep 18th, 2015

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