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NCERT Class XII Chemistry
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Question : 8 of 53
Marks: +1, -0
An antifreeze solution is prepared from 222.6 g of ethylene glycol,C2H4(OH)2\mathrm{C_2H_4(OH)_2} and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL−1^{-1}, then what shall be the molarity of the solution?
Solution:  
Mass of the solute, C2H4(OH)2=222.6 g\mathrm{C_2H_4(OH)_2}=222.6\text{ g}
Molar mass of solute, C2H4(OH)2=62 g mol−1\mathrm{C_2H_4(OH)_2}=62\text{ g}\ \mathrm{mol}^{-1}
∴\therefore Moles of the solute =222.6 g62 g mol−1=3.59= \frac{222.6\text{ g}}{62\text{ g}\ \mathrm{mol}^{-1}}=3.59
Mass of the solvent =200 g=0.200 kg=200\text{ g}=0.200\text{ kg}
Volume of solution =422.61.072 g mL−1=394.2 mL=0.3942 L= \frac{422.6}{1.072\text{ g}\ \mathrm{mL}^{-1}}=394.2\text{ mL}=0.3942\text{ L}
Molality =Number of moles of soluteMass of solvent in kg= \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} =3.590.2=17.95 m= \frac{3.59}{0.2}=17.95\text{ m}
Molarity =3.59 moles0.3942 L= \frac{3.59\text{ moles}}{0.3942\text{ L}} =9.11 mol L−1=9.11\ \mathrm{mol}\ \mathrm{L}^{-1}
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