### Molarity Problems#11 - 25

Problem #11: What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?

Solution:

12.0 M = 3.00 mol / x

x = 0.250 L

This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value:

0.250 L x (1000 mL / L) = 250. mL (note use of explicit decimal point to create three sig figs)

Problem #12: How many grams of Ca(OH)2 are needed to make 100.0 mL of 0.250 M solution?

Solution:

(0.250 mol L¯1) (0.100 L) = x / 74.0918 g mol¯1

x = (0.250 mol L¯1) (0.100 L) (74.0918 g mol¯1)

x = 1.85 g (to three sig figs)

Problem #13: What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution?

Solution:

(x) (0.0500 L) = 20.0 g / 97.9937 g mol¯1

(x) (0.0500 L) = 0.204094753 mol

x = 4.08 M

Problem #14: What weight (in grams) of KCl is there in 2.50 liters of 0.500 M KCl solution?

Solution:

(0.500 mol L¯1) (2.50 L) = x / 74.551 g mol¯1

x = 93.2 g

Problem #15: What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?

Solution:

(x) (0.2500 L) = 12.0 g / 39.9969 g mol¯1 <--- note use of L in set up, not mL x = 1.20 M

Problem #16: Determine the molarity of these solutions:

a) 4.67 moles of Li2SO3 dissolved to make 2.04 liters of solution.
b) 0.629 moles of Al2O3 to make 1.500 liters of solution.
c) 4.783 grams of Na2CO3 to make 10.00 liters of solution.
d) 0.897 grams of (NH4)2CO3 to make 250 mL of solution.
e) 0.0348 grams of PbCl2 to form 45.0 mL of solution.

Solution set-ups:

a) x = 4.67 mol / 2.04 L
b) x = 0.629 mol / 1.500 L
c) (x) (10.00 L) = 4.783 g / 106.0 g mol¯1
d) (x) (0.250 L) = 0.897 g / 96.09 g mol¯1
e) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯1

Problem #17: Determine the number of moles of solute to prepare these solutions:

a) 2.35 liters of a 2.00 M Cu(NO3)2 solution.
b) 16.00 mL of a 0.415-molar Pb(NO3)2 solution.
c) 3.00 L of a 0.500 M MgCO3 solution.
d) 6.20 L of a 3.76-molar Na2O solution.

Solution set-ups:

a) x = (2.00 mol L¯1) (2.35 L)
b) x = (0.415 mol L¯1) (0.01600 L)
c) x = (0.500 mol L¯1) (3.00 L)
d) x = (3.76 mol L¯1) (6.20 L)

Comment: the technique used is this:

MV = moles of solute

This particular variation of the molarity equation occurs quite a bit in certain parts of the acid base unit.

Problem #18: Determine the grams of solute to prepare these solutions:

a) 0.289 liters of a 0.00300 M Cu(NO3)2 solution.
b) 16.00 milliliters of a 5.90-molar Pb(NO3)2 solution.
c) 508 mL of a 2.75-molar NaF solution.
d) 6.20 L of a 3.76-molar Na2O solution.
e) 0.500 L of a 1.00 M KCl solution.
f) 4.35 L of a 3.50 M CaCl2 solution.

Solution set-ups:

a) (0.00300 mol L¯1) (0.289 L) = x / 187.56 g mol¯1
b) (5.90 mol L¯1) (0.01600 L) = x / 331.2 g mol¯1
c) (2.75 mol L¯1) (0.508 L) = x / 41.99 g mol¯1
d) (3.76 mol L¯1) (6.20 L) = x / 61.98 g mol¯1
e) (1.00 mol L¯1) (0.500 L) = x / 74.55 g mol¯1
f) (3.50 mol L¯1) (4.35 L) = x / 110.99 g mol¯1

Problem #19: Determine the final volume of these solutions:

a) 4.67 moles of Li2SO3 dissolved to make a 3.89 M solution.
b) 4.907 moles of Al2O3 to make a 0.500 M solution.
c) 0.783 grams of Na2CO3 to make a 0.348 M solution.
d) 8.97 grams of (NH4)2CO3 to make a 0.250-molar solution.
e) 48.00 grams of PbCl2 to form a 5.0-molar solution.

Solution set-ups:

a) x = 4.67 mol / 3.89 mol L¯1
b) x = 4.907 mol / 0.500 mol L¯1
c) (0.348 mol L¯1) (x) = 0.783 g / 105.99 g mol¯1
d) (0.250 mol L¯1) (x) = 8.97 g / 96.01 g mol¯1
e) (5.00 mol L¯1) (x) = 48.0 g / 278.1 g mol¯1

Problem #20: A student placed 11.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100. mL of the final solution?

Solution path #1:

1) Calculate molarity of first solution (produced by dissolving 11.0 g of glucose):

MV = grams / molar mass

(x) (0.100 L) = 11.0 g / 180.155 g/mol

x = 0.610585 mol/L (I'll carry a few guard digits.)

2) Calculate molarity of second solution (produced by diluting the first solution):

M1V1 = M2V2

(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)

x = 0.0244234 mol/L

3) Determine grams of glucose in 100. mL of second solution:

MV = grams / molar mass

(0.0244234 mol/L) (0.100 L) = x / 180.155 g/mol

x = 0.44 g

Solution path #2:

1) Calculate how much glucose you have in 20.0 mL of the first solution.

11.0 g is to 100. mL as x is to 20.0 mL

Cross-multiply and divide

100x = 11.0 times 20.0

x = 2.2 g

2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.
2.2 g is to 500. mL as x is to 100. mL

Cross-multiply and divide

500x = 2.2 times 100

x = 0.44 g

In 100 mL of the final solution are 0.44 g glucose.

Problem #21: Commercial bleach solution contains 5.25% (by mass) of NaClO in water. It has a density of 1.08 g/mL. Caculate the molarity of this solution. (Hints: assume you have 1.00 L of solution; molar mass of NaClO 74.4 g/mol)

Solution:

1) Determine mass of 1.00 L of solution:

(1.08 g/mL) (1000 mL) = 1080 g

2) Determine mass of NaClO in 1080 g of solution:

(1080 g) (0.0525) = 56.7 g

3) Determine moles of NaClO:

56.7 g / 74.4 g/mol = 0.762 mol

4) Determine molarity of solution:

0.762 mol / 1.00 L = 0.762 M

Problem #22: What is the molality (and molarity) of a 20.0% by mass hydrochloric acid solution? The density of the solution is 1.0980 g/mL.

Solution: 1) Determine moles of HCl in 100.0 g of 20.0% solution.

20.0 % by mass means 20.0 g of HCl in 100.0 g of solution.

20.0 g / 36.4609 g/mol = 0.548 mol

2) Determine molality:
0.548 mol / 0.100 kg = 5.48 m
3) Determine volume of 100.0 g of solution.
100.0 g / 1.0980 g/mL = 91.07468 mL
4) Determine molarity:
0.548 mol / 0.09107468 L = 6.02 m

Problem #23: 25.0 mL of 0.250 M KI, 25.0 mL of 0.100 K2SO4, and 15.0 mL of 0.100 M MgCl2 were mixed together in a beaker. What are the molar concentrations of I¯, Cl¯ and K+ in the beaker?

Solution:

1) Calculate the total volume of the mixed solutions:

25.0 mL + 25.0 mL + 15.0 mL = 65.0 mL

2) Concentration of iodide ion:

M1V1 = M2V2

(0.250 mol/L) (25.0 mL) = (x) (65.0 mL)

x = 0.09615 M

to three sig figs, 0.0962 M

3) Concentration of the chloride ion:

moles Cl¯ ---> (0.100 mol/L) (0.0150 L) (2 Cl¯ / 1 MgCl2) = 0.00300 mol

0.00300 mol / 0.065 L = 0.04615 M

to three sig figs, 0.0462 M

4) Concentration of the potassium ion:

moles K+ from KI ---> (0.250 mol/L) (0.0250 L) = 0.00625 mol
moles K+ from K2SO4 ---> (0.100 mol/L) (0.0250 L) (2 K+ / 1 K2SO4) = 0.00500 mol

0.00625 mol + 0.00500 mol = 0.01125 mol

0.01125 mol / 0.0650 L = 0.173 M

Problem #24: Calculate the total concentration of all the ions in each of the following solutions:

a. 3.25 M NaCl
b. 1.75 M Ca(BrO3)2
c. 12.1 g of (NH4)2SO3 in 615 mL in solution.

Solution:

1) the sodium chloride solution:

for every one NaCl that dissolves, two ions are produced (one Na+ and one Cl¯).

the total concentration of all ions is this:

(3.25 mol/L) times (2 total ions / 1 NaCl formula unit) = 6.50 M

2) the calcium bromate solution:

three total ions are produced for every one Ca(BrO3)2 that dissolves (one Ca2+ and two BrO3¯

the total concentration of all ions is this:

(1.75 mol/L) times (3 total ions / 1 Ca(BrO3)2 formula unit) = 5.25 M

3) the ammonium sulfite solution:

calculate the concentration of (NH4)2SO3:

(x) (0.615 L) = 12.1 g / 116.1392 g/mol

x = 0.169407 M <--- I'll carry some guard digits

calculate the concentration of all ions:

(NH4)2SO3 produces three ions for every one formula unit that dissolves.

0.169407 M times 3 = 0.508221 M

to three sig figs, 0.508 M

Problem #25: A solution of calcium bromide contains 20.0 g dm-3. What is the molarity of the solution with respect to calcium bromide and bromine ions.

Solution:

MV = mass / molar mass

(x) (1.00 L) = 20.0 g / 199.886 g/mol

x = 0.100 M

When CaBr2 ionizes, two bromide ions are released for every one CaBr2 that dissolves. That leads to this:

[Br-] = 0.200 M