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प्रश्न
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL−1, then what shall be the molarity of the solution?
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उत्तर १
Molar mass of ethylene glycol [C2H4(OH)2] = 2 × 12 + 6 × 1 + 2 × 16
= 62 g mol−1
Number of moles of ethylene glycol = `(222.6 "g")/(62 "g mol"^(-1))`
= 3.59 mol
Therefore, molality of the solution = `(3.59 "mol")/(0.200 "kg")`
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL−1
∴ Volume of the solution = `(422.6 "g")/(1.072 "g mL"^(-1))`
= 394.2 mL
= 0.3942 L
⇒ Molarity of the solution = `(3.59 "mol")/(0.3942 "L")`
= 9.11 M
उत्तर २
Molality (m) of the solution is given by
`m = w/(M') xx 1000/(w')`
In the present case, w = 222.6 g, M' = 62 g mol−1, w' = 200 g (Mol. mass of glycol = 62)
∴ `m = (w xx 1000)/(M' xx w')`
= `(222.6 xx 1000)/(62 xx 200)`
= 17.95 mol kg−1
`"Volume of solution" = "Mass"/"Density"`
= `(222.6 + 200)/(1.072)`
= 394.2 mL
Molarity is given by
`w = (M xx M' xx v)/1000`
∴ `M = (w xx 1000)/(M' xx v)`
= `(222.6 xx 1000)/(62 xx 394.2)`
= 9.11 mol L−1
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