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sergiy2304 [10]
3 years ago
10

3. From the following formula, calculate the quantities required to make 5 lb o hydrophilic ointment. Methylparaben 0.25 g Propy

lparaben 0.15 g Sodium lauryl sul ate 10 g Propylene glycol 120 g Stearyl alcohol 250 g White petrolatum 250 g Purif ed water, ad 1000 g
Chemistry
1 answer:
adell [148]3 years ago
7 0

The given question is incomplete. The complete question is as follows.

From the following formula, calculate the quantities required for each of the following ingredients to make 5 lb. (avoir) of hydrophilic ointment. All values to be in g, do not include units in your submitted answer; use no more than 3 decimal places. Convert using 1 lb = 454 g. Note: all values done by mass, including the water, so this is not a q.s situation (volume is irrelevant).

Methylparaben 0.25 g

Propylparaben 0.15 g

Sodium Lauryl Sulfate 10 g

Propylene Glycol 120 g

Stearyl Alcohol 240 g

White Petrolatum 250 g

Purified Water, to make 1000 g

Explanation:

For the given quantities, the sum is 620.4 g. Now, amount of purified water (to make 1000 g of ointment)  is as follows.

             1000 g - 620.4 g

           = 379.6 g

Now, to make 5 lb we will calculate the quantities as follows.

Formula weight = 1000 g

Therefore, desired quantity is as follows.

             5 lb = 5 x 454 g

                     = 2270 g

Hence, this means the factor is  calculated as follows.

       \frac{2270 g}{1000 g}

            = 2.27

Now, we will calculate the quantity of each ingredient using factor 2.27 as follows.

  • Methylparaben = 0.25 g x 2.27 = 0.5675 g
  • Propylparaben = 0.15 g x 2.27 = 0.3405 g
  • Sodium Lauryl Sulfate = 10 g x 2.27 = 22.7 g
  • Propylene Glycol = 120 g x 2.27 = 272.4 g
  • Stearyl Alcohol = 240 g x 2.27 = 544.8 g
  • White Petrolatum = 250 g x 2.27 = 567.5 g
  • Purified Water = 379.6 g x 2.27 = 861.692 g
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\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

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\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

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According to the Raoult's law,

p^o=X\times p_T

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p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

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and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

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