WS 8.9 Review Worksheet-Solutions Worksheet for 10th - 12th Grade ... - Free Printable

Let’s go through each question one by one. I’ll solve them step by step, check my work, and then give you the final answers at the end.

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Question 1: How much KCl can be dissolved in 100 g of water at 60°C?

We use the solubility curve for KCl (potassium chloride). At 60°C, find where the KCl line is on the graph.

Looking at standard solubility curves (and assuming this graph matches typical data):

→ At 60°C, KCl solubility ≈ 45 g per 100 g water

✔ Final Answer for Q1: 45 g

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Question 2: How much KNO₃ can be dissolved in 1000 g of water at 70°C?

First, find solubility of KNO₃ at 70°C from the graph.

Typical value: At 70°C, KNO₃ ≈ 135 g per 100 g water

But we have 1000 g water → that’s 10 times more.

So: 135 g × 10 = 1350 g

✔ Final Answer for Q2: 1350 g

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Question 3: How many grams of water are needed to dissolve 40.0 g of NaCl at 90°C?

Find solubility of NaCl at 90°C.

NaCl doesn’t change much with temperature — it’s almost flat.

At 90°C, NaCl solubility ≈ 39 g per 100 g water (very close to room temp)

Wait — let’s double-check: Actually, NaCl solubility is about 38–39 g/100g water even at high temps. Let’s say 39 g / 100 g water

We need to dissolve 40.0 g NaCl.

Set up proportion:

If 39 g NaCl needs 100 g water
Then 40.0 g NaCl needs ? g water

→ (40.0 / 39) × 100 ≈ 102.56 g water

Round to 3 significant figures? The given mass is 40.0 (3 sig figs), so answer should be 103 g

Wait — but let’s confirm actual solubility. Some sources say NaCl at 90°C is ~39.8 g/100g water. But since the graph likely shows ~39–40, and if we assume exactly 40 g/100g at 90°C (for simplicity in some graphs), then:

If solubility = 40 g / 100 g water → then 40.0 g NaCl needs exactly 100 g water

Hmm — this depends on the graph. Since the problem says “use the graph”, and if the graph shows NaCl at 90°C as approximately 40 g/100g, then:

✔ Final Answer for Q3: 100 g

*(Note: If your graph shows slightly different, adjust accordingly. But based on common textbook graphs, 40 g/100g at 90°C is acceptable.)*

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Question 4: What temperature would be required to get 40.0 g of KCl to dissolve in 100 g of water?

We need to find the temperature where KCl solubility = 40.0 g / 100 g water.

Look at KCl curve: Find where y = 40, read x (temperature).

From typical graph: KCl reaches 40 g/100g at around 50°C

Check: At 40°C → ~40 g? Wait no — at 40°C, KCl is about 40 g? Let me recall:

Standard values:
- 20°C: 34 g
- 40°C: 40 g
- 60°C: 45 g

Actually, at 40°C, KCl is about 40 g/100g water.

Yes — so to dissolve 40.0 g KCl in 100 g water, you need at least 40°C

✔ Final Answer for Q4: 40°C

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Question 5: Which substance can be dissolved to make a saturated solution of 40.0 g of KCl in 100 g of water?

This seems poorly worded. Probably means: “Which substance has a solubility of 40.0 g per 100 g water at some temperature?” OR maybe “At what temperature does KCl reach 40 g/100g?”

But looking back — Question 4 already asked for temperature for KCl.

Perhaps this is asking: “Which other substance (besides KCl) can also form a saturated solution with 40.0 g in 100 g water at some temperature?”

Looking at the graph (assumed):

- NaCl: always around 36–40 g — so yes, at higher temps, NaCl can reach 40 g/100g? Actually, NaCl max is ~39.8 g at 100°C — so not quite 40.0.

- KNO₃: at low temps, less than 40; at 25°C it’s ~38 g, at 30°C ~45 g — so between 25–30°C, KNO₃ hits 40 g.

- NH₄Cl: at 40°C, ~45 g; at 20°C, ~37 g — so around 25°C, NH₄Cl might be 40 g.

Actually, let’s think differently. Maybe the question is misphrased, and it’s asking: “What substance, when 40.0 g is added to 100 g water, will make a saturated solution at a certain temperature?” — but without specifying temp, it’s ambiguous.

Wait — perhaps it’s asking: “Which substance has a solubility of 40.0 g/100g water at the same temperature where KCl does?” — which is 40°C (from Q4).

At 40°C:

- KCl: 40 g
- NaCl: ~37 g → not saturated at 40 g
- KNO₃: ~65 g → way over
- NH₄Cl: ~45 g → over
- Ce₂(SO₄)₃: decreasing — very low

None match exactly except KCl itself.

Alternatively, maybe it’s a trick — only KCl makes a saturated solution with 40.0 g in 100 g water at 40°C.

But the question says “which substance” — implying another one.

Wait — look at NaCl: if we take 40.0 g NaCl in 100 g water at 90°C — from Q3, we saw that at 90°C, NaCl solubility is ~39–40 g, so 40.0 g would be *just* saturated or slightly supersaturated.

But technically, if solubility is 39.8 g, then 40.0 g is supersaturated — not saturated.

Perhaps the intended answer is NaCl, assuming the graph shows NaCl at 90°C as 40 g/100g.

Given ambiguity, and since Q3 used 40 g/100g for NaCl at 90°C, likely the graph assumes that.

So: At 90°C, 40.0 g NaCl in 100 g water = saturated.

Thus, NaCl can also make a saturated solution with 40.0 g in 100 g water — at 90°C.

✔ Final Answer for Q5: NaCl

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Question 6: 400 g of solution contains 14 mg of sugar.

Part A: What is % sugar in the solution?

% = (mass solute / mass solution) × 100%

Mass solute = 14 mg = 0.014 g
Mass solution = 400 g

% = (0.014 / 400) × 100 = 0.0035%

That’s tiny — but correct.

Part B: What is ppm in the solution?

ppm = parts per million = (mass solute / mass solution) × 1,000,000

= (0.014 g / 400 g) × 1,000,000 = 35 ppm

✔ Final Answers for Q6:
A: 0.0035%
B: 35 ppm

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Question 7: How many grams of HF would be in 15.0 kg of 30.0% HF solution?

30.0% by mass means 30.0 g HF per 100 g solution.

Total solution = 15.0 kg = 15,000 g

Mass HF = 30.0% of 15,000 g = 0.300 × 15,000 = 4,500 g

✔ Final Answer for Q7: 4500 g

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Question 8: How much 0.20% salt solution can be made using 10.7 g of salt?

0.20% means 0.20 g salt per 100 g solution.

So, set up:

0.20 g salt → 100 g solution
10.7 g salt → ? g solution

? = (10.7 / 0.20) × 100 = 53.5 × 100 = 5350 g

Or 5.35 kg

✔ Final Answer for Q8: 5350 g

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Question 9: What is the molarity of a solution containing 1.2 moles NaCl dissolved in 750 mL of NaCl solution?

Molarity (M) = moles / liters

Volume = 750 mL = 0.750 L

M = 1.2 mol / 0.750 L = 1.6 M

✔ Final Answer for Q9: 1.6 M

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Question 10: How many moles of sugar are needed to make 1.00 mL of 1.0 M sugar solution?

Molarity = moles / volume(L)

So, moles = M × V(L)

V = 1.00 mL = 0.00100 L

M = 1.0 mol/L

moles = 1.0 × 0.00100 = 0.0010 moles

Or 1.0 × 10⁻³ moles

✔ Final Answer for Q10: 0.0010 moles

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Now, compiling all final answers:

──────────────────────────────────────

Final Answer:
1. 45 g
2. 1350 g
3. 100 g
4. 40°C
5. NaCl
6. A: 0.0035%, B: 35 ppm
7. 4500 g
8. 5350 g
9. 1.6 M
10. 0.0010 moles