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A new solution is made by pouring 100 grams of water into
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23 Jul 2025, 22:46
23 Jul 2025, 22:46
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All solutions are composed of one of the compounds completely dissolved in water. Solubility refers to the ability of a substance to dissolve completely in water at a given temperature, and is expressed as a percent of compound per unit mass of solution. The mass of a solution is the sum of the mass of the water and the mass of the compound dissolved in the water.
A new solution is made by pouring 100 grams of water into a 160-gram solution that is 5% Compound X. At 50° C, what is the maximum number of grams of Compound X that can be added into the new solution so that all of the added Compound X dissolves?
a) 26
b) 20
c) 18
d) 10
e) 8
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A new solution is made by pouring 100 grams of water into
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03 Aug 2025, 08:38
03 Aug 2025, 08:38
Expert Reply
Given:
- Initial solution: 160 g total mass, $\(5 \)%$ Compound X .
- Mass of water added: 100 g .
- Temperature: $\(50^{\circ} \mathrm{C}\)$.
- We want to find the maximum grams of Compound $X$ that can be added to the new solution so that it all dissolves.
Step 1: Calculate initial amounts in the $\(\mathbf{1 6 0 ~ g}\)$ solution.
Compound $X$ mass in initial solution:
$$
\(5 % \times 160=0.05 \times 160=8 \text { grams }\)
$$
Water mass in initial solution:
$$
\(160-8=152 \text { grams }\)
$$
Step 2: Calculate total mass of water and compound after adding 100 g of water.
New water mass:
$$
\(152+100=252 \text { grams }\)
$$
Let the mass of the added Compound X to be $m$ grams.
Total Compound X mass after addition:
$$
\(8+m \text { grams }\)
$$
Total solution mass:
$$
\(252+8+m=260+m \text { grams }\)
$$
Step 3: Find maximum solubility of Compound X at $\(50^{\circ} \mathrm{C}\)$ from the graph.
From the graph in the attachment, at $\(50^{\circ} \mathrm{C}\)$ :
- The solubility of Compound X is approximately $\(8 \)%$ (percent of compound per unit mass of solution).
- Initial solution: 160 g total mass, $\(5 \)%$ Compound X .
- Mass of water added: 100 g .
- Temperature: $\(50^{\circ} \mathrm{C}\)$.
- We want to find the maximum grams of Compound $X$ that can be added to the new solution so that it all dissolves.
Step 1: Calculate initial amounts in the $\(\mathbf{1 6 0 ~ g}\)$ solution.
Compound $X$ mass in initial solution:
$$
\(5 % \times 160=0.05 \times 160=8 \text { grams }\)
$$
Water mass in initial solution:
$$
\(160-8=152 \text { grams }\)
$$
Step 2: Calculate total mass of water and compound after adding 100 g of water.
New water mass:
$$
\(152+100=252 \text { grams }\)
$$
Let the mass of the added Compound X to be $m$ grams.
Total Compound X mass after addition:
$$
\(8+m \text { grams }\)
$$
Total solution mass:
$$
\(252+8+m=260+m \text { grams }\)
$$
Step 3: Find maximum solubility of Compound X at $\(50^{\circ} \mathrm{C}\)$ from the graph.
From the graph in the attachment, at $\(50^{\circ} \mathrm{C}\)$ :
- The solubility of Compound X is approximately $\(8 \)%$ (percent of compound per unit mass of solution).
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