20 grams of Na2S04 is dissolved in 250 mL distilled water. What is the molarity

Question

20 grams of Na2S04 is dissolved in 250 mL distilled water. What is the molarity of the resulting solution?
50 mL of .1 M NaCl solution is mixed with 50 mL of .1 M MgCl2 solution. What is the molarity of Cl- ion in the resulting solution?

Explanation / Answer

First calculate the moles of NaCl using g NaCl. Molar mass of NaCl = 58.45 g Mole NaCl = 1.60 g NaCl ( 1 mole NaCl/ 58.45 g NaCl) = 0.0274 mole NaCl (answer to 3 sig.figs) Next calculate the moles of CaCl2 using M of CaCl2 and V (in L) of CaCl2. moles CaCl2 = (M) (L) = (0.100 M) (0.0500L) = (0.100 moles/L) (0.0500 L) = 0.00500 moles CaCl2 Find the concentration of Cl- ion from NaCl and CaCl2 separately and add the two to get mole Cl-. NaCl(aq) --> Na+(aq) + Cl-(aq) 0.0274 mole NaCl (1mole Cl-/1mole naCl) = 0.0274 mole Cl- CaCl2(aq) --> Ca^2+(aq) + 2 Cl-(aq) 0.00500 mole CaCl2 ( 2 mole Cl-/1 mole CaCl2) = 0.0100 mole Cl- Total concentration of Cl- = 0.0274 mole Cl- + 0.0100 mole Cl- = 0.0374 mole Cl- Molarity of Cl- = moles Cl-/L of solution = 0.0374 mole / 0.0500 L = 0.748 M

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