Question

An antifreezesolution prepared from 222.6g of ethylene glycol C2H2 (OH)2 and 200 g water.Calculate molarity of solution.If the density of the solution is 1.072g/ml Then what shall be the molarity of the solution?

Answer 1

Answer:

Calculation of Molality :

Mass of ethylene glycol = 222.6 (Given)

Molar mass of ethylene glycol [C2H4(OH)2]

= 2 X 12 + 6 x 1 + 2 x 16

= 62

Therefore moles of ethylene glycol

= 222.6g / 62 gmol-1

= 3.59 mol

Mass of water = 200g (Given)

Therefore molality of the solution is = (moles of ethylene glycol / mass of water) x 1000

= (3.59 / 200) x 1000

= 17.95 m

Calculation of Molarity:

Moles of ethylene glycol = 3.59 mol (already calculated)

Total Mass of solution = 200 + 222.6

= 422.6g

Volume of solution = mass / density volume

= 422.6 / 1.072

= 394.22 ml

now molarity of the solution is = (moles of ethylene glycol / volume of solution) x 1000

= (3.59 / 394.22) x 1000

= 9.11 M

Answer 2

$\huge\rm\underline\purple{Question :-}$

An antifreeze solution is prepared from 222.6 g of ethylene glycol and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g mL -1 what will be the molarity of the solutions?

$\huge\rm\underline\purple{Answer :-}$

$\blue{\underline\bold{Given,}}$

• Mass of the water = 200g
• Mass of ethylene glycol = 222.6g
• Density of the solution = 1.072

$\red{\underline\bold{To\:find,}}$

• Molality of the solution = ?
• Molarity of the solution = ?

$\green{\underline\bold{For\: finding\: molality,}}$

Moles of ethylene glycol = $\frac{222.6g}{62g/mol}$

= 3.59 mol

$\orange{\underline\bold{We\: know\: that,}}$

$\boxed{ \sf \purple{Molality =\frac{moles \:of \:ethylene\: glycol}{mass\: of \:water}× 100 }}$

= $\frac{3.59 mol}{200g}× 100$

= 17.95 m

$\blue{\underline\bold{Again,}}$

$\orange{\underline\bold{For\: finding\: molarity,}}$

Total mass of solution = 200 + 222.6

= 422.6 g

Volume of the solution = $\frac{422.6}{1.072}$

= $\frac{3.59 mol}{200g}× 100$

= 394.22 mL

$\boxed{ \sf \purple{Molarity =\frac{moles \:of \:ethylene\: glycol}{volume\: of \:solution}× 100 }}$

= $\frac{3.59 }{394.22}× 1000$

= 9.11 M

$\pink{\underline\bold{∴\: The\: required\:molality\:is\: 17.95\: m}}$

$\green{\underline\bold{and\: The\: required\:molarity\:is\: 9.11\: M}}$

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