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Mariana [72]
3 years ago
10

What is the solubility of NH4Cl at 50°C?

Chemistry
2 answers:
astra-53 [7]3 years ago
8 0

The correct answer is E. 50 grams

The solubility of Ammonium Chloride has an increasing trend as the temperature increases. Experiments show that the solubility at 323K is approximately 50.4 grams per 100 grams o water thus this value is close to choice D, 50 grams

maw [93]3 years ago
3 0

<u>Answer:</u> The correct answer is Option E.

<u>Explanation:</u>

Solubility is defined as the amount of substance (known as solute) which is dissolved in a unit volume of liquid substance (known as solvent) to produce a saturated solution at specific temperature and pressure. It is expressed in number of moles of solute present per 100 grams of solvent.

This property is directly dependent on temperature. As, the temperature increases, the solubility of the solute increases and vice-versa.

\text{Solubility}\propto \text{Temperature}

NH_4Cl is an ionic salt whose solubility increases as we increase the temperature.

The solubility of ammonium chloride at 50°C is 50 grams.

Hence, the correct answer is Option E.

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For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.
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Answer:

Trial     Number of moles

           

  1          0.001249mol

  2         0.001232mol

  3          0.001187 mol

Explanation:

To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.

Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.

<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.

In this case the solute is <em>NaOH</em>.

The formula is:

          Molarity=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}

Solve for the <em>number of moles:</em>

          \text{Number of moles}=Molarity\times Volume\text{ }in\text{ }liters

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.

Trial    mL           liters          Number of moles

           

1          12.49       0.01249        0.01249liters × 0.1000M = 0.001249mol

2         12.32      0.01232         0.01232liters × 0.1000M = 0.001232mol

3          11.87       0.01187         0.01187liters × 0.1000M = 0.001187 mol

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