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A mixture of ethyl iodide (b.p. 72.3) and water boils at 63.7 C. What weight of...

A mixture of ethyl iodide (b.p. 72.3) and water boils at 63.7 C. What weight of ethyl iodide would be carried over by 1 gm of steam during steam distillation
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The following reaction takes place during steam distillation of ethyl iodide,

CH_3CH_2I + H_2O \rightarrow CH_3CH_2OH + HI

Weight of water per gram of substance

=\frac{18 \times p_{H_2O } }{MW \ of\ substance \times (760 - p_{H_2O } )}

From the reaction, The following can be carried out:

1 mol ethyl iodide reacts with 1 mol of water, then 1 g of water has moles=

\frac{1\ g\ of \ H_2O\times1 mol\ H_2O}{18 g\ H_2O} = 0.055\ mol\ H_2O

So we need 0.055 mol of ethyl iodide i.e.

=\frac{(0.055 \ mol)(156 \ g ) }{1 mol} = 8.67 \ g

Weight of Ethyl iodide=8.67 g

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