Question

An antifreeze solution is prepared from 222.6 of ethylene glycol and 200g of water.Calculate the molality of the solution.

Answer 1

Answer:

Explanation:

Molar mass of ethylene glycol C2H4(OH)2 = 12 x 2 + 1 x 4 + (16+(1x 2))  

M2 = 24 + 4 + 34 = 62 g mol-1  

Mass of ethylene glycol, w2 = 22.6 g  

Mass of water, w1 = 200g  

Molality:

=222.6*1000/62*200

=17.95 m

Calculation of Molarity :

Total mass of antifreeze solution =222.6 +200 = 422.6 g

Volume of this solution V =  mass/density

                                          =422.6/1.072

                                          =394.22 ml

Molarity  :

=222.6*1000/62*394.22

=9.11M

Answer 2

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

An antifreeze solution is prepared from 222.6 of ethylene glycol and 200g of water. Calculate the molality of the solution.

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

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

• Mass of the water = 200g
• Mass of ethylene glycol = 222.6g

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

• Molality of the solution = ?

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

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

= 3.59 mol

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

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

= $\frac{3.59 mol}{200g}× 100$

= 17.95 m

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

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