JEE MAIN - Chemistry (2011 - No. 1)
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be
added to 4 kg of water to prevent it from freezing at −6oC will be :
[Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol = 62 g mol−1 )
204.30 g
400.00 g
304.60 g
804.32 g
Explanation
Given $${K_r} = 1.86\,K\,kg\,mo{l^{ - 1}}$$
$$\Delta {T_f} = 0 - \left( { - 6} \right) = {6^ \circ }C$$
As we know that
$$\Delta {T_f} = {K_f} \times \,\,molality$$
$$ = {{{K_f} \times 1000 \times mass\,\,of\,\,solute} \over {molar\,\,mass\,\,of\,\,solute\,\, \times \,\,mass\,\,of\,\,solvent\,\,in\,\,kg}}$$
Substituting given values in formula
$$6 = {{1.86 \times 1000 \times w} \over {62 \times 4}};$$
$$\,\,\,\,\,w = 0.8\,kg = 800\,gm$$
$$\Delta {T_f} = 0 - \left( { - 6} \right) = {6^ \circ }C$$
As we know that
$$\Delta {T_f} = {K_f} \times \,\,molality$$
$$ = {{{K_f} \times 1000 \times mass\,\,of\,\,solute} \over {molar\,\,mass\,\,of\,\,solute\,\, \times \,\,mass\,\,of\,\,solvent\,\,in\,\,kg}}$$
Substituting given values in formula
$$6 = {{1.86 \times 1000 \times w} \over {62 \times 4}};$$
$$\,\,\,\,\,w = 0.8\,kg = 800\,gm$$
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