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Questions  >>  CBSE XII  >>  Chemistry  >>  Solutions
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Q)

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

$\begin{array}{1 1}17.95m\\16.75m\\14.75m\\12.75m\end{array} $

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A)
Given :
Mass of solution (ethylene glycol)=222.6g
Molar mass of ethylene glycol $(C_2H_4(OH)_2)=62$
Therefore moles of ethylene glycol =$\large\frac{222.6}{62}=$$3.59$ mole
Mass of water (solvent) =200g
Molality =$\large\frac{\text{Moles of solute}}{\text{Mass of solvent (gm)}}$$\times 1000$
$\Rightarrow \large\frac{3.59}{200}$$\times 1000$
$\Rightarrow 17.95m$
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