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Q1.29:
Nalorphene ((C_{19}H_{21}NO_{3})), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of (1.5 \times 10^{-3} m) aqueous solution required for the above dose.

Solution :

Step 1: Identifying Given Variables and Constants

To determine the mass of the aqueous solution required, we extract the following parameters provided by the problem statement:

  • Mass of solute (Nalorphene, $C_{19}H_{21}NO_{3}$) required: $w_2 = 1.5 \text{ mg} = 1.5 \times 10^{-3} \text{ g}$
  • Molality of the solution ($m$): $1.5 \times 10^{-3} \text{ mol/kg}$
  • Molar Mass of Nalorphene ($M_2$):
    $M_2 = (19 \times 12.01) + (21 \times 1.008) + (1 \times 14.01) + (3 \times 16.00)$
    $M_2 = 228.19 + 21.168 + 14.01 + 48.00 = 311.368 \text{ g/mol} \approx 311.4 \text{ g/mol}$

Step 2: Calculating Moles of Solute

Using the relationship between mass, molar mass, and the number of moles ($n_2 = \frac{w_2}{M_2}$):

$n_2 = \frac{1.5 \times 10^{-3} \text{ g}}{311.4 \text{ g/mol}} \approx 4.817 \times 10^{-6} \text{ moles}$

Step 3: Determining the Mass of Solvent

Per the definition of molality ($m = \frac{n_2}{W_1 \text{ (in kg)}}$), where $W_1$ is the mass of the solvent:

$W_1 = \frac{n_2}{m} = \frac{4.817 \times 10^{-6} \text{ mol}}{1.5 \times 10^{-3} \text{ mol/kg}}$

$W_1 = 3.211 \times 10^{-3} \text{ kg} = 3.211 \text{ g}$

Step 4: Calculating Total Mass of the Solution

The total mass of the solution ($W_{total}$) is the sum of the mass of the solute ($w_2$) and the mass of the solvent ($W_1$):

$W_{total} = w_2 + W_1$ [By the Law of Conservation of Mass]

$W_{total} = (1.5 \times 10^{-3} \text{ g}) + 3.211 \text{ g}$

$W_{total} = 0.0015 \text{ g} + 3.211 \text{ g} = 3.2125 \text{ g}$

Visualization of Concentration Components

Solute (1.5 mg) + Solvent (3.211 g) Total Solution Mass: 3.2125 g

Final Solution: The mass of the aqueous solution required to provide a 1.5 mg dose is 3.2125 g.


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