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An antifreeze solution is prepared from 222.6 \(\mathrm{g}\) of ethylene glycol, \(\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)\) and \(200 \mathrm{~g}\) of water. Calculate the molality of the solution. If the density of the solution is \(1.072 \mathrm{~g}\) \(\mathrm{mL}^{-1}\), then what shall be the molarity of the solution?
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Mass of solute \(=222 \cdot 6 \mathrm{~g}\) Molar mass of solute, \(\mathrm{C}_2 \mathrm{H}_4(\mathrm{OH})_2\) \(=12 \times 2+4+2(12+1)=62 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\begin{aligned} \therefore \quad \text { Moles of solute } &=\frac{222 \cdot 6}{62}=3 \cdot 59 \\ \quad \text { Mass of solvent } &=200 \mathrm{~g} \end{aligned}\)
\(\therefore \quad\) Molality \(=\frac{3.59}{200} \times 1000=17.95 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Total mass of solution \(=\) Mass of solute \(+\) Mass of solvent \(=222.6+200=422 \cdot 6 \mathrm{~g}\)
Volume of solution \(=\frac{422 \cdot 6}{1 \cdot 072}=394.21 \mathrm{~mL}\).
\(\therefore \quad\) Molarity \(=\frac{3 \cdot 59}{394 \cdot 2} \times 1000=9 \cdot 1 \mathrm{~mol} \mathrm{~L}^{-1}\)
\(\begin{aligned} \therefore \quad \text { Moles of solute } &=\frac{222 \cdot 6}{62}=3 \cdot 59 \\ \quad \text { Mass of solvent } &=200 \mathrm{~g} \end{aligned}\)
\(\therefore \quad\) Molality \(=\frac{3.59}{200} \times 1000=17.95 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Total mass of solution \(=\) Mass of solute \(+\) Mass of solvent \(=222.6+200=422 \cdot 6 \mathrm{~g}\)
Volume of solution \(=\frac{422 \cdot 6}{1 \cdot 072}=394.21 \mathrm{~mL}\).
\(\therefore \quad\) Molarity \(=\frac{3 \cdot 59}{394 \cdot 2} \times 1000=9 \cdot 1 \mathrm{~mol} \mathrm{~L}^{-1}\)
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