Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.
Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.
Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
Answer:1.188
⋅
10
5
1.188 Times 10 to the fifth power
Explanation: