Question

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volumetric flask. You find that the solution has an osmotic pressure of 531 mm Hg at 25 °C. What is the composition of the mixture?"

Answer 1

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = $C \times 0.082 \times 298.15 K$

               C = 0.0285

This also means that,

  $\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = $2.85 \times 10^{-3 }$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

                              = $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

                   = $\frac{x}{369} + \frac{(1.00 - x)}{342}$

                  = $2.85 \times 10^{-3}$

             = x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g  and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.