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2 years ago

how much potassium bromide will form a saturated solution with 100 grams of water at 40 degrees celsius

Chemistry

1 answer:

Citrus2011 [14]2 years ago

4 0

Answer:

75.4 g.

Explanation:

Hello there!

In this case, since the solubility stands for the maximum amount of a solute that is completely dissolved in a specific amount of solvent, typically 100 g of water as the universal solvent; we need to take into account that the higher the temperature the higher the solubility for the most of the solids.

In such a way, since the solubility of potassium bromide at 20 °C is 65.3 g per 100 g of water, at 40 °C is 75.4 g per 100 g of water, proving the aforesaid. Therefore, 75.4 g of potassium bromide will form a saturated solution with 100 g of water at 40 °C.

Regards!

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Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?

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The question is incomplete, here is the complete question:

Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?

A. 5.10 mM glucose, 15.3 mM NaCl

B. 5.56 mM glucose, 16.7 mM NaCl

C. 0.556 mM glucose, 0.167 mM NaCl

D. 0.222 mM glucose, 1.11 mM NaCl

E. 0.556 mM glucose, 0.0667 mM NaCl

Answer: The concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For glucose:

Molarity of glucose solution = 0.100 M

Volume of solution = 25.0 mL

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of glucose}\times 1000}{25.0}\\\\\text{Moles of glucose}=\frac{(0.100\times 25.0)}{1000}=0.0025mol$

For NaCl:

Molarity of NaCl solution = 0.500 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.500M=\frac{\text{Moles of NaCl}\times 1000}{15.0}\\\\\text{Moles of NaCl}=\frac{(0.500\times 15.0)}{1000}=0.0075mol$

Total volume of solution = 25.0 + 15.0 + 450. = 490. mL

Now, calculating the concentration of glucose and NaCl in the solution by using equation 1:

Using conversion factor: 1 M = 1000 mM

For glucose:

Moles of glucose = 0.0025 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

$\text{Concentration of glucose}=\frac{0.0025\times 1000}{490}\\\\\text{Concentration of glucose}=0.0051M=5.10mM$

For NaCl:

Moles of NaCl = 0.0075 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

$\text{Concentration of NaCl}=\frac{0.0075\times 1000}{490}\\\\\text{Concentration of NaCl}=0.0153M=15.3mM$

Hence, the concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

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uranmaximum [27]

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Read 2 more answers