A mixture has 18 g water and 414 g ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture)

Question

A mixture has 18 g water and 414 g ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture) (A)  0.1  (B)  0.4  (C)  0.7  (D)  0.9

Tardigrade Admin · Accepted Answer

Mole fraction =( text moles of water / text moles of water + text moles of ethanol ) Mol. wt. of H 2 O =18 mol. wt. of C 2 H 5 OH =46 Mole fraction of water =(18 / 18/18 / 18+414 / 46) =(1/1+9)=(1/10)=0.1

Q. A mixture has $18 g$ water and $414 g$ ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture)

$0.1$

$0.4$

$0.7$

$0.9$

Mole fraction $= \frac{moles of water}{moles of water + moles of ethanol}$
Mol. wt. of $H_{2} O = 18 m o l$ .
wt. of $C_{2} H_{5} O H = 46$
Mole fraction of water $= \frac{18/18}{18/18 + 414/46}$
$= \frac{1}{1 + 9} = \frac{1}{10} = 0.1$

Q. A mixture has 18g water and 414g ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture)

0.1

0.4

0.7

0.9