Question

3. (hard) 85.0 g NaCl was dissolved in 1.50 L water. 765 mL of this solution was heated on a hotplate 488 mL of water was evaporated off. What is the resulting concentration of the sodium chloride die beleid en die gelution? and then 7b] = D.7652 488 m > 1,438L

Answer 1

mass NaCl = 85.0 g

moles NaCl = (mass NaCl) / (molar mass NaCl)

moles NaCl = (85.0 g) / (58.44 g/mol)

moles NaCl = 1.4545 mol

volume of solution = 1.50 L

concentration NaCl = (moles NaCl) / (volume of solution)

concentration NaCl = (1.4545 mol) / (1.50 L)

concentration NaCl = 0.970 M

concentration Cl^- = concentration NaCl

concentration Cl^- = 0.970 M

volume of solution taken = 765 mL

moles of Cl^- = (concentration Cl^-) * (volume of solution taken)

moles of Cl^- = (0.970 M) * (765 mL)

moles of Cl^- = 741.8 mmol

volume of solution remaining = (volume of solution taken) - (volume of water evaporated)

volume of solution remaining = (765 mL) - (488 mL)

volume of solution remaining = 277 mL

resulting concentration of Cl^- = (moles of Cl^-) / (volume of solution remaining)

resulting concentration of Cl^- = (741.8 mmol) / (277 mL)

resulting concentration of Cl^- = 2.68 M