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shepuryov [24]
2 years ago
14

at standard pressure, what is the temperature at which a saturated solution of NH4Cl has a concentration of 60g NH4CL/100 g H2O

Chemistry
1 answer:
Nataliya [291]2 years ago
5 0

Answer: Temperature = T, unknown

Saturated Solution, NH4Cl concentration = 60g/100g H2O = 0.6g NH4Cl/g H2O

Assume density of H2O = 1 g/ml

m = 0.6g NH4Cl/g H2O / 1 g/ml

m = 0.6g NH4Cl/ml

See the table of saturated solutions and identify the temperature at which the concentration of NH4Cl is 60g/100g H2O.

Explanation:  The line on the graph on reference table G indicates a saturated solution of NH4CL as a concentration of 60. g NH4 Cl/100. g H2O

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If the solubility of sodium chloride is 36 grams per 100 grams of water, which of the following solutions would be considered un
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Answer: If the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

Explanation:

A solution which contains the maximum amount of solute is called a saturated solution. Whereas a solution in which more amount of solute is able to dissolve is called an unsaturated solution.

Now, the number of moles present in 36 g of NaCl (molar mass = 58.4 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{36 g}{58.4 g/mol}\\= 0.616 mol

This shows that solubility of sodium chloride is 36 grams per 100 grams of water means a maximum of 0.616 mol of NaCl will dissolve in 100 mL of water.

So, a solution in which number of moles of NaCl are less than 0.616 mol per 100 mL then the solution formed will be an unsaturated solution.

  • As 5.8 moles of NaCl dissolved in 1 L (or 1000 mL) of water. So, moles present in 100 mL are calculated as follows.

Moles = \frac{5.8 mol}{1000 mL} \times 100 mL\\= 0.58 mol

  • Moles present in 100 mL of water for 3.25 moles of NaCl dissolved in 500 ml in water are as follows.

Moles = \frac{3.25 mol}{500 mL} \times 100 mL\\0.65 mol

  • Moles present in 100 mL of water for 1.85 moles of NaCl dissolved in 300 ml of water are as follows.

Moles = \frac{1.85 mol}{300 mL} \times 100 mL\\= 0.616 mol

Thus, we can conclude that if the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

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