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NCERT Class XII Chemistry
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Question : 29 of 53
Marks: +1, -0
Nalorphene (C19H21NO3)(\mathrm{C}_{19}\mathrm{H}_{21}\mathrm{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 1.5×10−31.5 \times 10^{-3} m aqueous solution required for the above dose.
Solution:  
. Given, m=1.5×10−3m = 1.5 \times 10^{-3}
Mass of solvent = 1 kg
Molar mass of C19H21NO3\mathrm{C}_{19}\mathrm{H}_{21}\mathrm{NO}_3 =19×12+21+14+48= 19 \times 12 + 21 + 14 + 48 =311 g mol−1= 311 \text{ g mol}^{-1}
∴ 1.5×10−31.5 \times 10^{-3} mole of C19H21NO3\mathrm{C}_{19}\mathrm{H}_{21}\mathrm{NO}_3 =1.5×10−3×311g= 1.5 \times 10^{-3} \times 311\mathrm{g} =0.467 g=467 mg= 0.467\text{ g} = 467\text{ mg}
∴ Mass of solution = 1000 g + 0.467 g = 1000.467 g
Thus, for 467 mg of nalorphene, solution required = 1000.467 g
For 1.5 mg of nalorphene, solution required = 1000.467467×1.5=3.21 g\frac{1000.467}{467} \times 1.5 = 3.21\text{ g}
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