Question

how many grams of ethylene glycol CH2OHCH2OH must be added to 37.8grams of water to give a freezing point of -0.150

Answer 1

28774, Chemistry, Other

how many grams of ethylene glycol CH2OHCH2OH must be added to 37.8 grams of water to give a freezing point of -0.150

Solution:

The extent of freezing point depression depends only on the solute concentration (molality concentration) that can be estimated by a simple linear relationship with the cryoscopic constant: $\Delta T_{\mathrm{F}} = K_{\mathrm{F}} \cdot C_{\mathrm{m}}$

Where, $\Delta T_{\mathrm{F}}$ - the freezing point depression, is defined as $T_{\mathrm{F (pure solvent)}} - T_{\mathrm{F (solution)}} = 0 - (-0.150) = 0.150$;

$K_{\mathrm{F}}$ is the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. For water, $K_{\mathrm{F}} = 1.853 \, \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}$;

$C_{\mathrm{m}}$ - molality concentration (mole per kg of solvent or math equation is: $C_{\mathrm{m}} = \frac{m}{M \cdot m(H_{2}O)}$).

From this equation the molality of solution is: $C_{\mathrm{m}} = \frac{\Delta T_{\mathrm{F}}}{K_{\mathrm{F}}} = \frac{0.150}{1.853} = 0.081 \, \mathrm{mol/kg}$.

From equation of $C_{\mathrm{m}}$ we calculate the mass of solute: $m = C_{\mathrm{m}} \cdot M \cdot m(H_{2}O)$.

The molar mass of $\mathrm{CH}_2\mathrm{OHCH}_2\mathrm{OH}$ is: $\mathrm{M(C_2H_6O_2)} = 12\cdot 2 + 1\cdot 6 + 16\cdot 2 = 62~\mathrm{g / mol}$

The mass of $\mathrm{CH}_2\mathrm{OHCH}_2\mathrm{OH}$ is: $m = 0.081\cdot 62\cdot \frac{37.8}{1000} = 0.19\mathrm{g}$

**Answer**: The mass of ethylene glycol is $0.19\mathrm{g}$.