Molarity Problems
Molarity Difination
The equations I will use are:
M = moles of solute / liters of solution
and
MV = grams / molar mass <--- The volume here MUST be in liters.
Typically, the solution is for the molarity (M). However, sometimes it is not, so be aware of that. A teacher might teach problems where the molarity is calculated but ask for the volume on a test question.
Note: Make sure you pay close attention to multiply and divide. For example, look at answer #8. Note that the 58.443 is in the denominator on the right side and you generate the final answer by doing 0.200 times 0.100 times 58.443.
Problem #1: Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?
Solution:
MV = grams / molar mass
(x) (1.00 L) = 28.0 g / 58.443 g mol¯1
x = 0.4790993 M
to three significant figures, 0.479 M
Problem #2: What is the molarity of 245.0 g of H2SO4 dissolved in 1.000 L of solution?
Solution:
MV = grams / molar mass
(x) (1.000 L) = 245.0 g / 98.0768 g mol¯1
x = 2.49804235 M
to four sig figs, 2.498 M
If the volume had been specified as 1.00 L (as it often is in problems like this), the answer would have been 2.50 M, NOT 2.5 M. You want three sig figs in the answer and 2.5 is only two SF.
Problem #3: What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?
Solution:
MV = grams / molar mass
(x) (0.4000 L) = 5.30 g / 105.988 g mol¯1
0.12501415 M
x = 0.125 M (to three sig figs)
Problem #4: What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?
Solution:
MV = grams / molar mass
(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1
(x) (0.7500 L) = 0.1250097 mol <--- threw in an extra step
x = 0.1666796 M
x = 0.167 M (to three SF)
Problem #5: How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?
Solution:
M = moles of solute / liters of solution
2.00 M = x / 10.0 L
x = 20.0 mol
Suppose the molarity was listed as 2.0 M (two sig figs). How to display the answer? Like this:
20. mol
Problem #6: How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?
Solution:
M = moles of solute / liters of solution
2.0 M = x / 0.0100 L <--- note the conversion of mL to L
x = 0.020 mol
Problem #7: How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?
Solution:
0.200 M = x / 0.1000 L
x = 0.0200 mol
Problem #8: What weight (in grams) of NaCl would be contained in problem #7?
Solution:
(0.200 mol L¯1) (0.100 L) = x / 58.443 g mol¯1 <--- this is the full set up
x = 1.17 g (to three SF)
You could have done this as well:
58.443 g/mol times 0.0200 mol <--- this is based on knowing the answer from problem #7
Problem #9: What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?
Solution:
(2.00 mol L¯1) (0.7500 L) = x / 98.0768 g mol¯1
x = (2.00 mol L¯1) (0.7500 L) (98.0768 g mol¯1)
x = 147.1152 g
to three sig figs, 147 g
Problem #10: What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?
Solution:
(18.0 mol L¯1) (x) = 2.45 g / 98.0768 g mol¯1
(18.0 mol L¯1) (x) = 0.0249804235 mol
x = 0.0013878 L
The above is the answer in liters. Multiplying the answer by 1000 provides the required mL value:
0.0013878 L times (1000 mL / L) = 1.39 mL (given to three sig figs)
Problem #11: What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?
Solution:
Problem #12: How many grams of Ca(OH)2 are needed to make 100.0 mL of 0.250 M solution?
Solution:
Problem #13: What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution?
Solution:
Problem #14: What weight (in grams) of KCl is there in 2.50 liters of 0.500 M KCl solution?
Solution:
Problem #15: What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?
Solution:
Problem #16: Determine the molarity of these solutions:
Problem #17: Determine the number of moles of solute to prepare these solutions:
Problem #18: Determine the grams of solute to prepare these solutions:
Problem #19: Determine the final volume of these solutions:
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):
Solution path #2:
1) Calculate how much glucose you have in 20.0 mL of the first solution.
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:
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.
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:
Problem #24: Calculate the total concentration of all the ions in each of the following solutions:
1) the sodium chloride solution:
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:
Problem #26: What is the concentration of each type of ion in solution after 23.69 mL of 3.611 M NaOH is added to 29.10 mL of 0.8921 M H2SO4? Assume that the final volume is the sum of the original volumes.
Solution:
The answer requires you to know how NaOH and H2SO4 react. Here is the chemical equation:
1) Calculate moles of NaOH and H2SO4:
Problem #27: Given 3.50 mL of sulfuric acid (98.0% w/w) calculate the number of mmols in the solution (density: 1.840 g/mL).
Solution:
Problem #28: Given 8.00 g of HBr calculate the volume (mL) of a 48.0% (w/w) solution. (MW HBr: 80.9119 g/mol, density: 1.49 g/mL). Then, calculate the molarity.
Solution:
Problem #29: A solution is made by dissolving 0.100 mol of NaCl in 4.90 mol of water. What is the mass % of NaCl?
Solution:
1) Convert moles to masses:
Problem #30: 2.00 L of HCl gas (measured at STP) is dissolved in water to give a total volume of 250. cm3 of solution. What is the molarity of this solution?
Solution using molar volume:
Bonus Problem: How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L of a solution that has a pH of 2.01?
Solution:
1) Get moles of hydrogen ion needed for the 18.0 L:
Molarity Difination
The equations I will use are:
M = moles of solute / liters of solution
and
MV = grams / molar mass <--- The volume here MUST be in liters.
Typically, the solution is for the molarity (M). However, sometimes it is not, so be aware of that. A teacher might teach problems where the molarity is calculated but ask for the volume on a test question.
Note: Make sure you pay close attention to multiply and divide. For example, look at answer #8. Note that the 58.443 is in the denominator on the right side and you generate the final answer by doing 0.200 times 0.100 times 58.443.
Problem #1: Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?
Solution:
MV = grams / molar mass
(x) (1.00 L) = 28.0 g / 58.443 g mol¯1
x = 0.4790993 M
to three significant figures, 0.479 M
Problem #2: What is the molarity of 245.0 g of H2SO4 dissolved in 1.000 L of solution?
Solution:
MV = grams / molar mass
(x) (1.000 L) = 245.0 g / 98.0768 g mol¯1
x = 2.49804235 M
to four sig figs, 2.498 M
If the volume had been specified as 1.00 L (as it often is in problems like this), the answer would have been 2.50 M, NOT 2.5 M. You want three sig figs in the answer and 2.5 is only two SF.
Problem #3: What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?
Solution:
MV = grams / molar mass
(x) (0.4000 L) = 5.30 g / 105.988 g mol¯1
0.12501415 M
x = 0.125 M (to three sig figs)
Problem #4: What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?
Solution:
MV = grams / molar mass
(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1
(x) (0.7500 L) = 0.1250097 mol <--- threw in an extra step
x = 0.1666796 M
x = 0.167 M (to three SF)
Problem #5: How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?
Solution:
M = moles of solute / liters of solution
2.00 M = x / 10.0 L
x = 20.0 mol
Suppose the molarity was listed as 2.0 M (two sig figs). How to display the answer? Like this:
20. mol
Problem #6: How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?
Solution:
M = moles of solute / liters of solution
2.0 M = x / 0.0100 L <--- note the conversion of mL to L
x = 0.020 mol
Problem #7: How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?
Solution:
0.200 M = x / 0.1000 L
x = 0.0200 mol
Problem #8: What weight (in grams) of NaCl would be contained in problem #7?
Solution:
(0.200 mol L¯1) (0.100 L) = x / 58.443 g mol¯1 <--- this is the full set up
x = 1.17 g (to three SF)
You could have done this as well:
58.443 g/mol times 0.0200 mol <--- this is based on knowing the answer from problem #7
Problem #9: What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?
Solution:
(2.00 mol L¯1) (0.7500 L) = x / 98.0768 g mol¯1
x = (2.00 mol L¯1) (0.7500 L) (98.0768 g mol¯1)
x = 147.1152 g
to three sig figs, 147 g
Problem #10: What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?
Solution:
(18.0 mol L¯1) (x) = 2.45 g / 98.0768 g mol¯1
(18.0 mol L¯1) (x) = 0.0249804235 mol
x = 0.0013878 L
The above is the answer in liters. Multiplying the answer by 1000 provides the required mL value:
0.0013878 L times (1000 mL / L) = 1.39 mL (given to three sig figs)
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 / xx = 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¯1x = (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¯1x = 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.Solution set-ups:
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.
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.Solution set-ups:
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.
a) x = (2.00 mol L¯1) (2.35 L)Comment: the technique used is this:
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)
MV = moles of soluteThis 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.Solution set-ups:
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.
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.Solution set-ups:
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.
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/mol2) Calculate molarity of second solution (produced by diluting the first solution):
x = 0.610585 mol/L (I'll carry a few guard digits.)
M1V1 = M2V2(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)3) Determine grams of glucose in 100. mL of second solution:
x = 0.0244234 mol/L
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 mLCross-multiply and divide2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.
100x = 11.0 times 20.0
x = 2.2 g
2.2 g is to 500. mL as x is to 100. mLCross-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 g2) Determine mass of NaClO in 1080 g of solution:
(1080 g) (0.0525) = 56.7 g3) Determine moles of NaClO:
56.7 g / 74.4 g/mol = 0.762 mol4) 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 mol2) Determine molality:
0.548 mol / 0.100 kg = 5.48 m3) Determine volume of 100.0 g of solution.
100.0 g / 1.0980 g/mL = 91.07468 mL4) 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 mL2) Concentration of iodide ion:
M1V1 = M2V2(0.250 mol/L) (25.0 mL) = (x) (65.0 mL)3) Concentration of the chloride ion:
x = 0.09615 M
to three sig figs, 0.0962 M
moles Cl¯ ---> (0.100 mol/L) (0.0150 L) (2 Cl¯ / 1 MgCl2) = 0.00300 mol0.00300 mol / 0.065 L = 0.04615 M4) Concentration of the potassium ion:
to three sig figs, 0.0462 M
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 mol0.00625 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 NaClSolution:
b. 1.75 M Ca(BrO3)2
c. 12.1 g of (NH4)2SO3 in 615 mL in 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:2) the calcium bromate solution:
(3.25 mol/L) times (2 total ions / 1 NaCl formula unit) = 6.50 M
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:3) the ammonium sulfite solution:
(1.75 mol/L) times (3 total ions / 1 Ca(BrO3)2 formula unit) = 5.25 M
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
Problem #26: What is the concentration of each type of ion in solution after 23.69 mL of 3.611 M NaOH is added to 29.10 mL of 0.8921 M H2SO4? Assume that the final volume is the sum of the original volumes.
Solution:
The answer requires you to know how NaOH and H2SO4 react. Here is the chemical equation:
H2SO4 + 2NaOH ---> Na2SO4 + 2H2OA key point is that two NaOH formula units are required for every one H2SO4
1) Calculate moles of NaOH and H2SO4:
moles NaOH ---> (3.611 mol/L) (0.02369 L) = 0.08554459 mol2) Determine how much NaOH remains after reacting with the H2SO4:
moles H2SO4 ---> (0.8921 mol/L) (0.02910 L) = 0.02596011 mol
0.02596011 mol x 2 = 0.05192022 mol <--- moles of NaOH that react0.08554459 mol - 0.05192022 mol = 0.03362437 mol <--- moles of NaOH that remain3) The above was required to determine the hydroxide ion concentration:
0.03362437 mol / 0.05279 L = 0.6369 M0.05279 L is the sum of the two solution volumes.4) Determine the sodium ion concentration:
0.08554459 mol / 0.05279 L = 1.620 MThe sole source of sodium ion is from the NaOH.5) Determine the sulfate ion concentration:
0.02596011 mol / 0.05279 L = 0.4918 MThe sole source of sulfate ion is from the H2SO4.6) Determine the hydrogen ion concentration:
[H+] [OH¯] = 1.000 x 10-14(x) (0.636945823) = 1.000 x 10-14
x = 1.570 x 10-14 M
Problem #27: Given 3.50 mL of sulfuric acid (98.0% w/w) calculate the number of mmols in the solution (density: 1.840 g/mL).
Solution:
3.50 mL times 1.840 g/mL = 6.44 g <--- mass of the 3.50 mL6.44 g times 0.980 = 6.3112 g <--- mass of H2SO4 in the solution
6.3112 g / 98.0768 g/mol = 0.06434957 mol
0.06434957 mol times (1000 mmol / 1 mol) = 64.3 mmol (to three sig figs)
Problem #28: Given 8.00 g of HBr calculate the volume (mL) of a 48.0% (w/w) solution. (MW HBr: 80.9119 g/mol, density: 1.49 g/mL). Then, calculate the molarity.
Solution:
8.00 g divided by 0.48 = 16.6667 g <--- total mass of the solution in which the HBr is 48% by mass16.6667 g divided by 1.49 g/mL = 11.18568 mL
to three sig figs, the volume of the solution is 11.2 mL
For the molarity, determine the moles of HBr:
8.00 g / 80.9119 g/mol = 0.098873 mol
0.098873 mol / 0.01118568 L = 8.84 M
Problem #29: A solution is made by dissolving 0.100 mol of NaCl in 4.90 mol of water. What is the mass % of NaCl?
Solution:
1) Convert moles to masses:
NaCl ---> 0.100 mol times 58.443 g/mol = 5.8443 gH2O ---> 4.90 mol times 18.015 g/mol = 88.2735 g2) Calculate mass percent of NaCl:
[5.8443 g / (5.8443 g + 88.2735 g)] * 100 = 6.21% (to three sig figs)
Problem #30: 2.00 L of HCl gas (measured at STP) is dissolved in water to give a total volume of 250. cm3 of solution. What is the molarity of this solution?
Solution using molar volume:
2.00 L divided by 22.414 L/mol = 0.0892299 mol of HCl0.0892299 mol / 0.250 L = 0.357 M (to three sig figs)Solution using Ideal Gas Law:
PV = nRT(1.00 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (273.15 K)If the volume of HCl gas was not at STP, you must use PV = nRT to calculate the moles. You cannot use molar volume since it is only true at STP.
n = 0.0892272 mol
0.0892272 mol / 0.250 L = 0.357 M (to three sig figs)
Bonus Problem: How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L of a solution that has a pH of 2.01?
Solution:
1) Get moles of hydrogen ion needed for the 18.0 L:
[H+] = 10-pH = 10-2.01 = 0.0097724 M0.0097724 mol/L) (18.0 L) = 0.1759032 mol of HCl required2) Determine molarity of 36.0% HCl:
Remember, HCl is a strong acid, dissociating 100% in solution
Assume 100. g of solution present.36.0 g of that is HCl3) Volume of 11.6508 M acid needed to deliver 0.1759032 mol:
100. g / 1.18 g/mL = 84.745763 mL
Use MV = mass / molar mass
(x) (0.084745763 L) = 36.0 g / 36.4609 g/mol
x = 11.6508 M
0.1759032 mol / 11.6508 mol/L = 0.01509795 L15.1 mL (to three sig figs)
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