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Drupady [299]
2 years ago
12

A student carried out a simple distillation on a compound known to boil at 124 oc and they reported a boiling point of 116-117 o

analysis of the compound showed it was pure and calibration of the thermometer indicated that it was accurate. What procedural error might the student have made in setting up the distillation?

Chemistry
1 answer:
chubhunter [2.5K]2 years ago
4 0

Answer :

The correct answer is due to incorrect position of thermometer .

Distillation  is process of separating two volatile liquids on the basis of their boiling points .In involves the following set up ( shown in image )

Since the Boiling point is found   is 116-117 C which is near to 124 C , also the purity was checked and thermometer was calibrated .

So the only error could be possible is position of thermometer .

The position of  bulb of thermometer is very important to get an accurate reading of boiling point .The vapor so formed should touch the side walls of bulb as soon as they are formed .

May be  the bulb of thermometer was   kept far from the vapors formed by liquid so,  s<u>ome of  the vapor before reaching to  bulb start condensing and their Temperature decreases </u>.  

Hence , the  temperature shown by thermometer was decrease than actual boiling point  due to incorrect position of thermometer.


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Iron fluoride (FeF2) dissociates according to the following equation:
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Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

Kps = \frac{[product]^x}{[reagent]^y}

Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.

Analyzing the equation, we see that for 1 mol of Fe^{+2} is necessary 2 mols of F^-, so if we call "x" the molar concentration of Fe^2, for F^- we will have 2x, so:

Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L

So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:

Fe = 55.8 g/mol

F = 19 g/mol

FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol

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[tex]S = 8.4x10^{-3}x93.8

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4 0
3 years ago
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
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Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

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