Question

6. Calculate the following, and make sure to show your work, including units and substance on every numerical value:

              a. What is the molarity of a solution that is made of 2.768 g of Pb(NO3)2 in 150 mL of H2O?

              b. How many grams of calcium chloride would it take to make 2.7L of a .10M solution?

c. What is the concentration of a solution made by diluting 100 mL of 3.0 M HCl with 300 mL water?

              c. How many milliliters of 6.0 M HNO3 is needed to make 20 L of .35 M HNO3?

              d. If 35 mL of 1.5 M lead nitrate reacts in a double replacement reaction with 40 mL of 2.0 M potassium iodide, how many grams of lead (II) iodide can be produced? (Hint: You will need a balanced equation first).

Answer 1

a. What is the molarity of a solution that is made of 2.768 g of Pb(NO3)2 in 150 mL of H2O?

M = mol/V

mol = mass/MW

M = (mass/MW)/V

M = (2.768)/(331.2098)/(0.150) = 0.05571 M

b. How many grams of calcium chloride would it take to make 2.7L of a .10M solution?

mol = M*V = 0.10*2.7 = 0.27 mol

mol of CaCl2 = 0.27 mol

mass = mol*MW = 0.27*110.98 = 29.9646 g

c. How many milliliters of 6.0 M HNO3 is needed to make 20 L of .35 M HNO3?

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

M1V1 = M2V2

20*0.35 = 6*V

V = 20*0.35 /6

V = 1.1666 Liter of stock solution required

d. If 35 mL of 1.5 M lead nitrate reacts in a double replacement reaction with 40 mL of 2.0 M potassium iodide, how many grams of lead (II) iodide can be produced? (Hint: You will need a balanced equation first).

mmol of Lead = MV = 1.5*35 = 52.5

mmol of Iodide = MV = 40*2 = 80

ratio is 1:2 so there is exces Pb

then, max amount of PbI2 = 80/2 = 40 mmol

mass = mol*MW = (40*10^-3)*461.00894 = 18.440 g