An antifreeze mixture consists of $40$ % ethylene glycol $(C_{2} H_{6} O_{2})$ by weight in aqueous solution. If the density of this solution is $1.05$ g/mL, what is the molar concentration?

6.77 M

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6.45 M

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0.017 M

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16.9 M

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Solution

The correct option is A 6.77 M
Take $100$ g aqueous solution, then, Mass of $(C_{2} H_{6} O_{2})$ = $40$ g.
Molar mass of $(C_{2} H_{6} O_{2})$ = $62$ g.
No. of moles of $(C_{2} H_{6} O_{2})$ in $100$ g solution = $\frac{40}{62}$ = $0.645$ moles.
Given, density = $1.05$ g/ml.
Volume of $100$ g solution = $\frac{M a s s}{d e n s i t y}$ = $\frac{100}{1.05}$ = $95.23$ ml = $0.09523$ L.
So, Molar Concentration = $\frac{M o l e s}{V o l u m e o f S o l u t i o n}$ = $\frac{0.645 M o l}{0.09523 L}$ = $6.77 M$

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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?

QUESTION 2.8

An antifreeze solution is prepared from 222.6 g of ethylene glycol $(C_{2} H_{6} O_{2})$ and 200 g of water. Calculate the molality of the solution. If the density of the solution is $1.072 g m L^{- 1}$ , then what shall be the molarity of the solution?