Question

1. A solution is created by dissolving 12.5 grams of ammonium chloride in enough water to make 215 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

2. When thinking about the amount of solute present in a solution, chemists report the concentration or molarity of the solution. Molarity is calculated as moles of solute per liter of solution. What is the molarity of the solution described above?

3. To carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume of the solution described above will you need to complete the reaction without any leftover NH₄Cl?

Answer 1

sol". Ansi1: How many moles of NH 4cl are present in Here in weight of NH 4 is 12.5 m mw of Nuad is 53.49 gmlmwe. Moles of NH 4d is weight of NHuel mw of Nuud 12.5 gm 53.49 ghi nel Moles of Nihad = 0.233 mol Molarity of above solution. Molurity CM) 2 am) lit - mW . M : wt. of solute who of solute x Vol.of soll = 1000 x Wt. of solute wt. of solute vol. of solnimmi 1000 X 12.5 12.5x215 (M = 4.65M Ansi-3 Volume of soll need to complete reaction. - Here 0.233 mol Nhad used in 215 ml solvent. above solution =) Is we add 0.0500 moles of Nhac in so, we need vol. of solm is... = 0.0500 x 215 - 0.233 = 46:13 min -) we need 46.13 ml + 215 ml = 261 13 ml total of soll to complete the reaction. [ Ans: 261-13 mi)