What volume of phenytoin suspension 30 mg/5 mL is required to be added to a suitable diluent to obtain 150 mL phenytoin suspensi
1 answer:
Explanation:
The given data is as follows.
Concentration of phenytoin suspension stock = 30 mg/5mL
Concentration of phenytoin required = 20 mg/5 mL
Volume of phenytoin required = 150 mL
Volume of phenytoin suspension stock required for dilution will be calculated as follows.
= 100 mL
Thus, we can conclude that the volume of phenytoin is 100 mL .
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Answer:
a) 1,6%
b) 64,775 g/mol
c) 3,6×10⁻² M
d) 2,3×10⁻³ g/mL
Explanation:
a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:
- Helium: 0,25 moles ×
= 1 g of Helium
- Argon: 0,25 moles ×
= 10 g of Argon
- Krypton: 0,25 moles ×
= 20,95 g of krypton
- Xenon: 0,25 moles ×
= 32,825 g of Xenon
Total grams: 1g+10g+20,85g+30,825g= 62,675 g
Mass fraction of helium: × 100 = <em>1,6%</em>
<em />
<em>The mass fraction of Helium is 1,6%</em>
<em />
<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:
= 64,775 g/mol
c) The molar concentration is possible to know ussing ideal gas law, thus:
= M
Where:
P is pressure: 150 kPa
R is gas constant: 8,3145
T is temperature: 500 K
And M is molar concentration. Replacing:
M = 3,6×10⁻² M
d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:
3,6×10⁻² ×
×
=
2,3×10⁻³ g/mL
I hope it helps!