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Archy [21]
3 years ago
15

The vapor pressure of benzene is 100.0 mmHg at 26.1°C. Calculate the vapor pressure of a solution containing 26.6 g of camphor (

C10H16O) dissolved in 94.0 g of benzene. (Camphor is a low-volatility solid.)
Chemistry
1 answer:
aliya0001 [1]3 years ago
6 0

Hey there!:

Molar mass benzene = 78.11 g/mol

find number of moles of benzene :

Moles of benzene =  mass / molar mass of benzene

Moles of benzene =  94.0 / 78.11

Moles of benzene = 1.203 moles

Molar mass camphor =  152.23 g/mol

Moles of camphor = 26.6 / 152.23

Moles of camphor = 0.1747 moles

Therefore , vapor pressure of solution is :

mole fraction of benzene * vapor pressure of solution

= 1.203 / ( 1.203 + 0.1747 ) * 100.0

=  ( 1.230 / 1.3777 ) * 100.0

= 0.8927 * 100.0

=> 89.27 mmHg

Hope this helps !

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
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Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Best regards.

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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
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Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

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