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bija089 [108]
3 years ago
5

What volume of phenytoin suspension 30 mg/5 mL is required to be added to a suitable diluent to obtain 150 mL phenytoin suspensi

on 20 mg/5 mL? 100
Chemistry
1 answer:
nydimaria [60]3 years ago
7 0

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.

                20 \times \frac{150}{30}

                  = 100 mL

Thus, we can conclude that the volume of phenytoin is 100 mL .

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a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
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Answer:

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xb= 0.24

xt=0.76

molar composition for vapor

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For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

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