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?​

Answer 1

$\huge{\bf{\color{turquoise}{Question:-}}}$

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?

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

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

• 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 = ?

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

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

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

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

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

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

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

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Answer 2

Upar wali meri bandi hai yaad rakhna xD