6. Calculate the following, and make sure to show your work, including units and

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).

Explanation / Answer

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

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