An antifreeze solution is prepared from 222.6 g of ethylene glycol, C2H4(OH)2 and 200 g of water. If the density of the solution is 1.072 g mL-1, the molality of the solution will be

Question

An antifreeze solution is prepared from 222.6 g of ethylene glycol, C2H4(OH)2 and 200 g of water. If the density of the solution is 1.072 g mL-1, the molality of the solution will be (A) 20.9 m (B) 17.95 m (C) 25.5 m (D) 35.7 m

Tardigrade Admin · Accepted Answer

Mass of the solute, C2H4(OH)2 = 222.6 g Molar mass of solute C2H4(OH)2 = 62 g mol -1 ∴ Moles of the solute = (222.6 g/62 g mol-1) = 3.59 Mass of the solvent = 200 g = 0.200 kg Volume of solution = (422.6/1.072 g mL-1) = 394.2 mL = 0.3942 L Molality = ( textNumber of moles of solute/ textMass of solvent in kg) = (3.59/0.2) = 17.95 m

Q. An antifreeze solution is prepared from $222.6 g$ of ethylene glycol, $C_{2} H_{4} (O H)_{2}$ and $200 g$ of water. If the density of the solution is $1.072 g m L^{- 1}$ , the molality of the solution will be

$20.9 m$

$17.95 m$

$25.5 m$

$35.7 m$

Q. An antifreeze solution is prepared from 222.6g of ethylene glycol, C2​H4​(OH)2​ and 200g of water. If the density of the solution is 1.072gmL−1, the molality of the solution will be

20.9m

17.95m

25.5m

35.7m

Q. An antifreeze solution is prepared from $222.6 g$ of ethylene glycol, $C_{2} H_{4} (O H)_{2}$ and $200 g$ of water. If the density of the solution is $1.072 g m L^{- 1}$ , the molality of the solution will be

$20.9 m$

$17.95 m$

$25.5 m$

$35.7 m$