JEE MAIN - Chemistry (2012 - No. 1)
Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of
ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8oC ?
72 g
93 g
39 g
27 g
Explanation
$$\Delta {T_f} = i \times {K_f} \times m$$
Given $$\Delta {T_f} = 2.8,{K_f} = 1.86\,K\,kg\,mo{l^{ - 1}}\,\,i = 1$$
(ethylene glygol is a non- electrolyte)
wt. of solvent $$=1$$ $$kg;$$
Let of wt of solute $$=x$$
Mol. wt of ethyllene glycol $$=62$$
$$2.8 = 1 \times 1.86 \times {x \over {62 \times 1}}$$
or $$\,\,\,\,\,\,x = {{2.8 \times 62} \over {1.86}} = 93gm$$
Given $$\Delta {T_f} = 2.8,{K_f} = 1.86\,K\,kg\,mo{l^{ - 1}}\,\,i = 1$$
(ethylene glygol is a non- electrolyte)
wt. of solvent $$=1$$ $$kg;$$
Let of wt of solute $$=x$$
Mol. wt of ethyllene glycol $$=62$$
$$2.8 = 1 \times 1.86 \times {x \over {62 \times 1}}$$
or $$\,\,\,\,\,\,x = {{2.8 \times 62} \over {1.86}} = 93gm$$
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