If you use 623 g of water, what mass (in g) of ethylene glycol should you use?

Question

You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.111.

Answer 1

Mole fraction = moles solute / moles solute + moles solvent

Molar mass H2O = 18 g /mol
623 / 18 = 34.6

0.111 = moles solute / 34.6 + moles solute

0.111 ( 34.6 + moles solute ) = moles solute

moles solute = 4.3

molar mass HOCH2CH2OH = 62 g/mol
62 g /mol x 4.3 mol = 267 g

Answer 2

Do you mean a concentration of 0.111 mol/dm^3?
no of moles of ethylene required = concentration x volume
= 0.111 x 623/1000
= 0.069153 mol
mass = no. of moles x Mr
= 0.069153 x (6 x 1.0 + 2 x 12.0 + 2 x 16.0)
= 4.287486 g

Answer 3

This is a mole fraction problem. Mole fraction = moles of solute/moles of solute + solvent. 623g H2O x 1molH2O/18gH2O= 34.61molH2O

0.111 = moles solute/34.61mol + mol solute

3.84 + molsolute = mole solute
3.84 + 1 = 4.84 = mole solute
(since you don't know the moles of solute, just use the # 1)

convert 4.84molethlene glycol into grams. You should know how to do this. Just take 4.84 and times by the mass of HOCH2CH2OH. Hope this helped :)

Answer 4

I only 002 10.0 points For the endothermic reaction H2 (g) + I2 (g) 2 HI(g) 1 which of ... 008 10.0 points 30.2 g of glycerine (C3 H8 O3 ) are dissolved in 150 g of water. ... Solution I with a boiling point of 100.32 C Explanation: First calculate the .... (1 kg water) = 167 g (CH2 OH)2 8 030 10.0 points What mass of ethylene glycol ..

Answer 5

1 mol of EG weighs = 62 g/mol
623 g/H20 = .623 L
do you have a density of EG = rho:

Mass of EG = .623L * .111 Mol/L * 62 g/Mol = 4.287 g

In a beaker put 4.287 g of EG
fill the beaker to .623 L
stir well and verify still at .623 L.

discard the remaining water (if any).

Answer 6

moles/total moles
(623 g H2O)/(18.0 g/mol) = 34.6 mol
x/(x+34.6 mol) = .111
x/.111 = x + 34.6
x = 4.25 moles HOCH2CH2OH * 62.0 g/mol = 263.5 g --> 264 g in significant figures, which a previous answer seemed to drop in an intermediate step of calculating the moles, hence the rounding error.

Answer 7

4.32 moles