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trapecia [35]
3 years ago
13

You need to purify 2.0 grams of an impure sample of Acetanilide. The sample is contaminated with aniline. After the purification

is complete you isolate 0.8 grams of acetanilide and record a melting point range of 108-110 °C. Complete the following calculations and show your work.
a. Calculate the minimum amount of distilled water you would use to complete the recrystallization.
b. How much acetanilide will still be dissolved in solution even after the sample is cooled to 0 °C?
c. Calculate the % recovery and the % error for the melting point.
d. Why is the percent recovery less than 100%? Describe multiple sources for loss of sample.

Chemistry
1 answer:
marusya05 [52]3 years ago
3 0

Answer:

Following are the answer to this question:

Explanation:

In the given question an attachment file is missing, that is attached. please find the attached file, and the following are the description of the given points:

a. At 100 degrees in 100 mL 5 g is dissolved.  

For, it required:

\to 2g = 100 \times \frac{2}{5}

         = 40 \ \ ml \ of \ water.  

b. At 0 degrees 100 mL dissolve in 0.3 g.  

So, the dissolve:

\to 40 \ ml= 0.3\times \frac{40}{100}

               = 0.12g.

After refrigeration 0.12 g will still be dissolved.  

c. After dissolving and freezing, precipitation can occur which would still be impure if the cooling is instantaneous. The added solvent was also too hard to recrystallize. The solvent was placed below its place of reservation.  

d. Recovery percentage:

\to \frac{0.8}{2}\times100

\to 40 \ \%

The melting point of acetanilide:

\to 114.3^{\circ}.

Found=109(medium)  

Melting point error percentages:

= \frac{114.3-109}{114.3}\\\\=4.63 \ \%

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4. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0˚C in a calorimeter,
myrzilka [38]

Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

Volume of 1.00 M Na2SO4 = 1.00 L

enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL

Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

Q= m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

7 0
3 years ago
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