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Likurg_2 [28]
3 years ago
6

The freezing point of diet soda is higher than the freezing point of regular soda, but lower than 0 degrees celcius, the freezin

g point of pure water. explain why both diet and regular soda freeze at temperatures lower thaan 0 degree celcius
Chemistry
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

Explanation:

Both Diet Soda and regular soda contain sweeteners.

When a solute is dissolved in solution, the solution undergoes *freezing point depression* it freezing point reduces. The magnitude of freezing point depression is directly proportional to the amount of solute in a solution.

Since soda Both regular or diet soda contains more solute than water , their freezing point is will consequently be lower than water

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volume = 20 cm³

density = mass / volume

therefore:

D = m / V

D = 5 / 20

D = 0.25 g/cm³
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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
Whitepunk [10]
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
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k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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