Question

A---Which of the following solutions of strong electrolytes contains the largest total number of ions? Show your calculations and explain your reasoning.

100.0 mL of 0.100 M sodium hydroxide

50.00 mL of 0.200 M barium chloride

75.0 mL of 0.150 M sodium phosphate

B--35.0 mL of lead(II) nitrate solution reacts with excess sodium iodide solution to yield 0.628 g of precipitate.

--Write a balanced equation for this reaction, and identify the precipitate. Indicate states of matter with parentheses and (aq) or (s).
--How many moles of precipitate formed?
--How many moles of lead nitrate were in the original solution?
--What is the concentration of the lead nitrate solution in grams per mL?

C---A gas mixture, with a total pressure of 300.0 torr, consists of equal masses of Ne (atomic weight 20.18) and Ar (atomic weight 39.95). What is the partial pressure of Ar, in torr?

Answer 1

A---Which of the following solutions of strong electrolytes contains the largest total number of ions? Show your calculations and explain your reasoning.

Solution :- using the molarity and volume we can calculate the moles of the each solute and then depending on the number of ions produced by each compound we can calculate the total number of ions.

100.0 mL of 0.100 M sodium hydroxide

Moles total moles = (0.100 M * 0.100 L)*2 = 0.02 mol ions

50.00 mL of 0.200 M barium chloride

Moles of ions = (0.200 M * 0.050 L) * 3 = 0.03 mol ions

75.0 mL of 0.150 M sodium phosphate

Moles of ions = (0.150 M * 0.075 L)*4 = 0.045 mol ions

So the Sodium phosphate solution gives the highest number ions

B--35.0 mL of lead(II) nitrate solution reacts with excess sodium iodide solution to yield 0.628 g of precipitate.

Write a balanced equation for this reaction, and identify the precipitate. Indicate states of matter with parentheses and (aq) or (s).
Balanced reaction equation is as follows

Pb(NO3)2(aq) + 2NaI(aq) -------- > PbI2(s) + 2naNO3(aq)

--How many moles of precipitate formed?

Moles of PbI2 = mass / molar mass

                       = 0.628 g / 461.01 g per mol

                       = 0.001362 mol PbI2

How many moles of lead nitrate were in the original solution?
solution :- mole ratio of the PbI2 to Pb(NO3)2 is 1 : 1

Therefore moles of Pb(NO3)2 present in the solution = 0.001362 mol Pb(NO3)2

--What is the concentration of the lead nitrate solution in grams per mL?

Solution :- mass of Pb(NO3)2 = moles * molar mass

                                                = 0.001362 mol * 331.2 g per mol

                                                = 0.451 g Pb(NO3)2

Concentration in g/ml = 0.451 g / 35.0 ml = 0.0129 g/ml

C---A gas mixture, with a total pressure of 300.0 torr, consists of equal masses of Ne (atomic weight 20.18) and Ar (atomic weight 39.95). What is the partial pressure of Ar, in torr?

Solution :- equal mass of the Ne and Ar

Lets assume we have 10 g of each gas

Then moles of Ar = 10 g / 39.95 g per mol = 0.250 mol

Moles of Ne = 10 g / 20.18 g per mol = 0.496 mol Ne

Now lets calculate the partial pressure of the Ar

Partial pressure of Ar= (moles of Ar/ total moles )* total pressure

                                    = (0.250 mol / (0.250 +0.496 ))*300 torr

                                    = 100 torr

So the partial pressure of Ar = 100 torr