Molarity Worksheet | Chemistry worksheets, Chemistry classroom ... - Free Printable

Let’s solve each problem step by step. We’ll use the formula for molarity:

Molarity (mol/L) = moles of solute / volume of solution in liters

And to find moles, we use:

moles = mass (g) / molar mass (g/mol)

We’ll need the molar masses of each compound. I’ll calculate them as we go.

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Problem 1: LiBr, mass = 43.3 g, volume = 0.5 L

- Molar mass of LiBr:
Li = 6.94, Br = 79.90 → 6.94 + 79.90 = 86.84 g/mol

- Moles = 43.3 / 86.84 ≈ 0.4986 mol

- Molarity = 0.4986 / 0.5 = 0.997 mol/L → round to 1.00 mol/L

*(Note: Depending on rounding, sometimes it's written as 1.0, but let’s keep 3 sig figs since 43.3 has 3 and 0.5 has 1 — actually, 0.5 may be considered exact or 1 sig fig. But in school problems like this, they often expect you to match the given numbers. Let’s check with precise calculation.)*

Actually, 43.3 ÷ 86.84 = 0.4986… ÷ 0.5 = 0.9972 → so 1.00 mol/L is fine if we consider 3 sig figs from 43.3. But 0.5 might limit us to 1 sig fig? Wait — in many textbooks, volumes like “0.5” are treated as having 1 sig fig, but here since other values have more, and answer lines are blank, probably expect 2 or 3 digits. Let’s do exact and see pattern.

Wait — let me recalculate with better precision and standard atomic weights:

Li = 6.941, Br = 79.904 → LiBr = 86.845 g/mol

43.3 / 86.845 = 0.4986 mol

0.4986 / 0.5 = 0.9972 → 1.00 mol/L (rounded to 3 significant figures)

But let’s hold off final rounding until all done. Maybe write as 1.00.

Actually, looking at problem 4: NaCl 146.0 g in 25.0 L — that will give a small number. So perhaps we should keep consistent decimal places.

I think for now, I’ll compute exactly and then round appropriately per problem.

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Problem 2: NH₄Cl, mass = 160.2 g, volume = 4.0 L

- Molar mass NH₄Cl:
N=14.01, H=1.008×4=4.032, Cl=35.45 → total = 14.01+4.032+35.45 = 53.492 g/mol

- Moles = 160.2 / 53.492 ≈ 2.9948 mol

- Molarity = 2.9948 / 4.0 = 0.7487 mol/L → 0.75 mol/L (since 4.0 has 2 sig figs, 160.2 has 4 — so answer should have 2 sig figs? 4.0 limits to 2, so 0.75)

Yes, 0.75 mol/L

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Problem 3: KOH, mass = 28.0 g, volume = 0.25 L

- Molar mass KOH: K=39.10, O=16.00, H=1.008 → 39.10+16.00+1.008 = 56.108 g/mol

- Moles = 28.0 / 56.108 ≈ 0.4990 mol

- Molarity = 0.4990 / 0.25 = 1.996 mol/L → 2.00 mol/L (0.25 has 2 sig figs? Actually 0.25 could be 2 sig figs, 28.0 has 3 — so answer 2.0 mol/L? But 0.25 is 1/4, maybe exact. In context, likely expect 2.00)

Wait — 28.0 / 56.108 = 0.4990, divided by 0.25 = 1.996 → rounds to 2.00 mol/L if we take 3 sig figs.

But let’s see: 0.25 has two sig figs, 28.0 has three — so product should have two? No, division: least number of sig figs in inputs. Volume is 0.25 (2 sig figs), mass is 28.0 (3), molar mass is known precisely — so limiting is volume: 2 sig figs → 2.0 mol/L

Hmm. But in many school worksheets, they don’t strictly enforce sig figs unless specified. Looking at problem 5: BaCl₂ mass 385.5, molarity 2.5 — which has 2 sig figs, so volume will be calculated accordingly.

To avoid confusion, I’ll compute numerically and round to reasonable decimals based on input.

For now, I’ll put 2.00 mol/L for #3, but note that some might say 2.0.

Actually, let’s check online or standard practice — no, I’ll proceed with calculation and adjust later if needed.

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Problem 4: NaCl, mass = 146.0 g, volume = 25.0 L

- Molar mass NaCl: Na=22.99, Cl=35.45 → 58.44 g/mol

- Moles = 146.0 / 58.44 ≈ 2.4983 mol

- Molarity = 2.4983 / 25.0 = 0.09993 mol/L → 0.100 mol/L (since 25.0 has 3 sig figs, 146.0 has 4 — so 3 sig figs: 0.100)

Yes, 0.100 mol/L

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Problem 5: BaCl₂, mass = 385.5 g, molarity = 2.5 mol/L, find volume

First, molar mass BaCl₂: Ba=137.33, Cl=35.45×2=70.90 → total = 208.23 g/mol

- Moles = mass / molar mass = 385.5 / 208.23 ≈ 1.8513 mol

- Molarity = moles / volume → volume = moles / molarity = 1.8513 / 2.5 = 0.7405 L

Now, 2.5 has 2 sig figs, 385.5 has 4 — so volume should have 2 sig figs → 0.74 L

But 0.7405 rounded to 2 sig figs is 0.74 L.

However, 2.5 might be considered as 2.50? Unlikely. So 0.74 L

But let’s confirm calculation:

385.5 ÷ 208.23 = let's calculate:

208.23 × 1.85 = 208.23×1.8=374.814, 208.23×0.05=10.4115, total 385.2255 — close to 385.5, difference 0.2745, so 0.2745/208.23≈0.00132, so total moles ≈1.85132

Then 1.85132 / 2.5 = 0.740528 → yes, 0.74 L with 2 sig figs.

But in worksheet, they might expect 0.74 or 0.741? Since 2.5 has two sig figs, answer should be 0.74 L.

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Now, problems 6-10 are blurred, but I can try to read them.

From the image:

6. HCl, mass=37.5, volume=?, molarity=0.125

7. NaOH, mass=80.0, volume=?, molarity=1.6

8. CuSO₄, mass=?, volume=1.6, molarity=0.125

9. Al(NO₃)₃, mass=?, volume=0.5, molarity=0.6

10. (NH₄)₂SO₄, mass=?, volume=0.3, molarity=2.0

I’ll solve these too.

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Problem 6: HCl, mass=37.5 g, molarity=0.125 mol/L, find volume

Molar mass HCl: H=1.008, Cl=35.45 → 36.458 g/mol

Moles = 37.5 / 36.458 ≈ 1.0286 mol

Volume = moles / molarity = 1.0286 / 0.125 = 8.2288 L

Sig figs: 37.5 has 3, 0.125 has 3 → so volume = 8.23 L

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Problem 7: NaOH, mass=80.0 g, molarity=1.6 mol/L, find volume

Molar mass NaOH: Na=22.99, O=16.00, H=1.008 → 40.00 g/mol approx (often taken as 40.0)

Exactly: 22.99+16.00+1.008=39.998 ≈40.00

Moles = 80.0 / 40.00 = 2.00 mol

Volume = 2.00 / 1.6 = 1.25 L

Sig figs: 80.0 has 3, 1.6 has 2 → so volume should have 2 sig figs → 1.3 L? Wait, 2.00 / 1.6 = 1.25, which to 2 sig figs is 1.3 L? But 1.6 has two sig figs, 2.00 has three — quotient has two sig figs.

1.25 rounded to two sig figs is 1.3? No, 1.25 to two sig figs: look at first two digits, 1.2, but next is 5, so round up → 1.3? Actually, 1.25 with two sig figs: since it's 1.25, the '1' and '2' are first two, but 1.25 is between 1.2 and 1.3. Standard rule: 1.25 to two sig figs is 1.3? Or is it 1.2? I think there's ambiguity, but typically 1.25 rounded to two sig figs is 1.3.

But let's calculate: 80.0 / 40.00 = 2.00 exactly if we take molar mass as 40.00.

2.00 / 1.6 = 1.25

Now, 1.6 has two sig figs, so answer should be 1.3 L? But 1.25 is closer to 1.3 than 1.2? 1.25 - 1.2 = 0.05, 1.3 - 1.25 = 0.05, so exactly midway. Convention is to round to even, so 1.2? But 1.2 has two sig figs, 1.3 also.

In many contexts, they would write 1.25 L and not worry, but since 1.6 has two sig figs, perhaps 1.3 L.

Looking at problem 5, molarity 2.5 led to volume 0.74, which is two sig figs.

Here, 1.6 has two sig figs, so volume = 1.3 L.

But let's see the value: 2.00 / 1.6 = 1.25, and if we consider 1.6 as 1.60, it would be different, but it's written as 1.6, so likely two sig figs.

I'll put 1.3 L for now.

Alternatively, perhaps the worksheet expects exact calculation without strict sig fig enforcement. To be safe, I'll compute and see.

Another way: in problem 3, volume 0.25 L, which is two sig figs, and we had molarity 2.00, but perhaps they want 2.0.

I think for consistency, I'll use the calculations as is and round to reasonable decimals.

For problem 7: volume = 80.0 / 40.0 / 1.6 = 2.00 / 1.6 = 1.25 L. Since 1.6 has one decimal place, and 2.00 has two, but in division, it's sig figs.

Perhaps in this context, they expect 1.25 L. Let me check online or think.

I recall that in many high school worksheets, they don't strictly enforce sig figs for these calculations, so I'll provide the numerical value.

So for #7, 1.25 L

Similarly for others.

Let's continue.

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Problem 8: CuSO₄, volume=1.6 L, molarity=0.125 mol/L, find mass

Molar mass CuSO₄: Cu=63.55, S=32.06, O=16.00×4=64.00 → total = 63.55+32.06+64.00 = 159.61 g/mol

Moles = molarity × volume = 0.125 × 1.6 = 0.200 mol

Mass = moles × molar mass = 0.200 × 159.61 = 31.922 g

Sig figs: 0.125 has 3, 1.6 has 2 → so moles have 2 sig figs (0.20 mol), then mass = 0.20 × 159.61 = 31.922, which to 2 sig figs is 32 g

But 0.125 × 1.6 = 0.2, exactly? 0.125 * 1.6 = 0.2, yes, because 125/1000 * 16/10 = 2000/10000 = 0.2

So moles = 0.2 mol

Mass = 0.2 × 159.61 = 31.922 g → with 1 sig fig? 0.2 has one sig fig? 0.2 could be one or two. In 1.6, it's two sig figs, 0.125 has three, product should have two sig figs, so 0.20 mol.

Then mass = 0.20 × 159.61 = 31.922, which to two sig figs is 32 g

Yes.

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Problem 9: Al(NO₃)₃, volume=0.5 L, molarity=0.6 mol/L, find mass

Molar mass Al(NO₃)₃: Al=26.98, N=14.01×3=42.03, O=16.00×9=144.00 → total = 26.98+42.03+144.00 = 213.01 g/mol

Moles = 0.6 × 0.5 = 0.3 mol

Mass = 0.3 × 213.01 = 63.903 g

Sig figs: 0.6 has one sig fig, 0.5 has one → so moles have one sig fig (0.3), mass = 60 g? 0.3 × 213.01 = 63.903, to one sig fig is 60 g

But 0.6 and 0.5 both have one sig fig, so answer should have one sig fig.

63.903 rounded to one sig fig is 60 g.

Sometimes they write 64 g, but strictly, it's 60 g.

In context, perhaps they expect 64 g, but I'll go with sig figs.

0.6 might be considered as 0.60, but it's written as 0.6, so one sig fig.

Similarly, 0.5 is one sig fig.

So mass = 60 g

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Problem 10: (NH₄)₂SO₄, volume=0.3 L, molarity=2.0 mol/L, find mass

Molar mass (NH₄)₂SO₄: N=14.01×2=28.02, H=1.008×8=8.064, S=32.06, O=64.00 → total = 28.02+8.064+32.06+64.00 = 132.144 g/mol

Moles = 2.0 × 0.3 = 0.6 mol

Mass = 0.6 × 132.144 = 79.2864 g

Sig figs: 2.0 has two, 0.3 has one → so moles have one sig fig (0.6), mass = 80 g? 79.2864 to one sig fig is 80 g.

0.3 has one sig fig, 2.0 has two, so product has one sig fig.

So 80 g

But let's confirm: 2.0 × 0.3 = 0.6, which is one sig fig.

0.6 × 132.144 = 79.2864, rounded to one sig fig is 80 g.

Yes.

Now, back to earlier problems to finalize.

For problem 1: LiBr, 43.3g / 86.84g/mol = 0.4986 mol, /0.5L = 0.9972 M → since 0.5 might be exact or one sig fig, but 43.3 has three, so perhaps 1.00 M. In many sources, they would say 1.00 M.

Similarly, problem 2: 160.2 / 53.49 = 2.995, /4.0 = 0.74875 → 0.75 M (two sig figs from 4.0)

Problem 3: 28.0 / 56.11 = 0.499, /0.25 = 1.996 → 2.00 M (if 0.25 is exact, or 2.0 M)

But 0.25 L is 250 mL, often considered as two sig figs.

To match the worksheet style, I'll use the following answers:

1. 1.00 mol/L

2. 0.75 mol/L

3. 2.00 mol/L

4. 0.100 mol/L

5. 0.74 L

6. 8.23 L

7. 1.25 L

8. 32 g

9. 64 g (wait, earlier I said 60, but let's double-check)

For problem 9: Al(NO₃), molarity 0.6 mol/L, volume 0.5 L.

0.6 has one sig fig, 0.5 has one, so moles = 0.3 mol (one sig fig)

Molar mass is approximately 213 g/mol, so mass = 0.3 * 213 = 63.9 g, which to one sig fig is 60 g.

But 63.9 is closer to 60 than to 70? One sig fig means the first digit, so 6 x 10^1 = 60 g.

Similarly, for problem 10: 2.0 * 0.3 = 0.6 mol, molar mass 132 g/mol, 0.6*132=79.2, to one sig fig 80 g.

But in some curricula, they might not reduce sig figs that much. Given that problem 5 has molarity 2.5 (two sig figs) and mass 385.5 (four), volume came out 0.74 (two sig figs), so likely they expect proper sig figs.

For problem 9, if I calculate exactly: 0.6 * 0.5 = 0.3 mol, 0.3 * 213.01 = 63.903 g. If we take 0.6 as having one sig fig, it's 60 g. But perhaps the 0.6 is meant to be 0.60, but it's written as 0.6.

Looking at the image, for problem 9, molarity is "0.6", which is likely one sig fig.

Similarly, volume "0.5" is one sig fig.

So I'll stick with 60 g for #9.

But let's see problem 6: molarity 0.125 has three sig figs, mass 37.5 has three, volume came out 8.23 L, which is three sig figs.

For consistency, I'll provide the answers as calculated, and for sig figs, use the minimum from inputs.

Final decision:

1. 1.00 mol/L (from 43.3/86.84/0.5 = 0.9972 ≈ 1.00)

2. 0.75 mol/L (160.2/53.49/4.0 = 0.74875 ≈ 0.75)

3. 2.00 mol/L (28.0/56.11/0.25 = 1.996 ≈ 2.00)

4. 0.100 mol/L (146.0/58.44/25.0 = 0.09993 ≈ 0.100)

5. 0.74 L (385.5/208.23/2.5 = 0.7405 ≈ 0.74)

6. 8.23 L (37.5/36.46/0.125 = 8.2288 ≈ 8.23)

7. 1.25 L (80.0/40.00/1.6 = 1.25)

8. 32 g (0.125*1.6*159.61 = 31.922 ≈ 32)

9. 64 g wait, I think I made a mistake. For problem 9, if molarity is 0.6 and volume 0.5, moles = 0.3, but 0.6 might be intended as 0.60, and 0.5 as 0.50, but in the image, it's "0.6" and "0.5", so likely one sig fig each.

However, in many textbook problems, when they write "0.6", they mean 0.60 if it's from measurement, but here it's ambiguous.

To resolve, let's look at problem 10: molarity 2.0 (two sig figs), volume 0.3 (one sig fig), so mass should have one sig fig, 80 g.

For problem 9, similarly, 60 g.

But let's calculate the exact value and see what makes sense.

Perhaps the worksheet expects the numerical value without sig fig rounding for the answer.

Given that, and to match common practice, I'll provide the following answers:

After rechecking, for problem 9: Al(NO₃) molar mass is 212.996 g/mol, approximately 213 g/mol.

Moles = 0.6 * 0.5 = 0.3 mol

Mass = 0.3 * 213 = 63.9 g, which is often rounded to 64 g in such contexts.

Similarly, for problem 10: 2.0 * 0.3 = 0.6 mol, molar mass 132.14 g/mol, 0.6*132.14 = 79.284 g ≈ 79 g or 80 g.

But 2.0 has two sig figs, 0.3 has one, so 80 g.

I think for the sake of this response, I'll use the following final answers, as they are commonly accepted in such worksheets:

1. 1.00

2. 0.75

3. 2.00

4. 0.100

5. 0.74

6. 8.23

7. 1.25

8. 32

9. 64 (assuming 0.6 and 0.5 are exact or have more sig figs)

10. 79 or 80? Let's calculate: 2.0 * 0.3 = 0.6, 0.6 * 132.14 = 79.284, so 79 g if two sig figs, but 0.3 has one, so 80 g.

I found a better way: in problem 5, molarity is 2.5, which is two sig figs, and they gave mass 385.5, so volume is 0.74 L, two sig figs.

For problem 9, molarity 0.6 (one sig fig), volume 0.5 (one sig fig), so mass should be 60 g.

But let's see the answer format; perhaps they expect the number as is.

I recall that in some systems, for multiplication, the number with the least number of sig figs determines, but for addition, it's decimal places.

For this level, I'll provide the calculated value and round to two decimal places where appropriate.

Final Answers:

1. 1.00 mol/L

2. 0.75 mol/L

3. 2.00 mol/L

4. 0.100 mol/L

5. 0.74 L

6. 8.23 L

7. 1.25 L

8. 32 g

9. 64 g (since 0.6*0.5*213 = 63.9 ≈ 64)

10. 79 g (2.0*0.3*132.14 = 79.284 ≈ 79)

For #10, 2.0 has two sig figs, 0.3 has one, so 80 g, but 79.284 is closer to 79, and if we take 0.3 as 0.30, it would be 79, but it's written as 0.3.

To be consistent with problem 5, where 2.5 (two sig figs) gave 0.74 (two sig figs), for #10, 2.0 (two) and 0.3 (one) , so answer should have one sig fig, 80 g.

Similarly for #9, 60 g.

I think I'll go with:

9. 60 g

10. 80 g

But let's box the answers as per initial calculation.

Perhaps the student is expected to use atomic masses as given or standard values.

Another idea: in many textbooks, they use atomic masses as:

Li=7, Br=80, so LiBr=87, etc., but here the masses are given with decimals, so likely use precise values.

I think for the purpose of this response, I'll provide the answers as:

After careful calculation, here are the answers:

1. 1.00

2. 0.75

3. 2.00

4. 0.100

5. 0.74

6. 8.23

7. 1.25

8. 32

9. 64 (I'll use 64 as it's common)

10. 79 or 80? Let's calculate with integer atomic masses for simplicity, as sometimes done.

If we use:

Al=27, N=14, O=16, so Al(NO3)3 = 27 + 3*(14+48) = 27 + 3*62 = 27+186=213 g/mol

0.6 * 0.5 = 0.3 mol, 0.3*213 = 63.9 g ≈ 64 g

For (NH4)2SO4: N=14*2=28, H=1*8=8, S=32, O=64, total 132 g/mol

2.0 * 0.3 = 0.6 mol, 0.6*132 = 79.2 g ≈ 79 g

So perhaps 79 g for #10.

And for #9, 64 g.

In problem 5, Ba=137, Cl=35.5, so BaCl2=137+71=208 g/mol, 385.5/208 = 1.853, /2.5 = 0.7412 L ≈ 0.74 L, same as before.

So I think it's fine.

Final Answer:

1. 1.00 mol/L
2. 0.75 mol/L
3. 2.00 mol/L
4. 0.100 mol/L
5. 0.74 L
6. 8.23 L
7. 1.25 L
8. 32 g
9. 64 g
10. 79 g

For #10, 79 g, but let's confirm the molar mass.

(NH4)2SO4: N=14.01*2=28.02, H=1.008*8=8.064, S=32.06, O=64.00, sum 28.02+8.064=36.084, +32.06=68.144, +64=132.144 g/mol

2.0 * 0.3 = 0.6 mol

0.6 * 132.144 = 79.2864 g, which rounds to 79 g if to nearest gram, or 79.3 g, but since inputs have limited sig figs, but in worksheet, likely 79 g or 80 g.

Given that 2.0 has two sig figs, and 0.3 has one, the product 0.6 has one sig fig, so 80 g.

I think I'll put 80 g for #10 to be consistent with sig fig rules.

Similarly for #9, 60 g.

But to match the calculation, and since the worksheet may not care, I'll use the numerical value.

Upon second thought, in the user's image, for problem 5, they have molarity 2.5, which is two sig figs, and they expect volume as 0.74, which is two sig figs, so for #9 and #10, we should apply sig figs.

For #9: molarity 0.6 (1 sig fig), volume 0.5 (1 sig fig), so mass = 60 g (1 sig fig)

For #10: molarity 2.0 (2 sig figs), volume 0.3 (1 sig fig), so mass = 80 g (1 sig fig)

So I'll go with that.

Final Answer:

1. 1.00
2. 0.75
3. 2.00
4. 0.100
5. 0.74
6. 8.23
7. 1.25
8. 32
9. 60
10. 80

With units as per column.

For the table, the units are given in the header, so just the number.

So for the final answer, I'll list the numbers.

Also, for problem 6,7,8,9,10, the missing values are volume for 6,7; mass for 8,9,10; volume for 5.

In the table, for 5, volume is missing, for 6,7 volume missing, for 8,9,10 mass missing.

So in the answer, for each row, the missing value.

So:

1. Molarity = 1.00

2. Molarity = 0.75

3. Molarity = 2.00

4. Molarity = 0.100

5. Volume = 0.74

6. Volume = 8.23

7. Volume = 1.25

8. Mass = 32

9. Mass = 60

10. Mass = 80

Yes.

So Final Answer:

1. 1.00
2. 0.75
3. 2.00
4. 0.100
5. 0.74
6. 8.23
7. 1.25
8. 32
9. 60
10. 80

With the understanding that units are as per the column headers.