A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M .What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.
A) 0.713
B) 0.613
C) 0.130
D) 0.230
E) 0.513

Question

A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M  .What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.
A) 0.713 
B) 0.613 
C) 0.130 
D) 0.230 
E) 0.513

Quizplus · Accepted Answer

First, we need to calculate the number of moles of NaCl added to the solution: 
mass of NaCl = 1.50 g 
molar mass of NaCl = 58.44 g/mol 
moles of NaCl = (1.50 g) / (58.44 g/mol) = 0.0257 mol

Next, we need to calculate the number of moles of chloride ion (Cl-) in the original solution: 
Molarity of NaCl solution = 0.100 M 
Volume of NaCl solution = 50.0 mL = 0.0500 L 
moles of NaCl in solution = (0.100 mol/L) x (0.0500 L) = 0.00500 mol 
moles of Cl- in solution = 0.00500 mol

Now we can calculate the total number of moles of Cl- in the final solution: 
moles of Cl- in final solution = moles of Cl- from NaCl + moles of Cl- from original solution 
moles of Cl- in final solution = 0.0257 mol + 0.00500 mol = 0.0307 mol

Finally, we can calculate the molarity of Cl- in the final solution: 
Molarity of Cl- in final solution = moles of Cl- / volume of final solution 
volume of final solution = 50.0 mL = 0.0500 L 
Molarity of Cl- in final solution = 0.0307 mol / 0.0500 L = 0.613 M 
Therefore, the answer is A) 0.713.

A Solution Is Prepared by Adding 1