'X’ grams of water is mixed in 69 grams of ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of ‘X’ in grams

Question

'X’ grams of water is mixed in 69 grams of ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of ‘X’ in grams (A) 54 (B) 36 (C) 180 (D) 18

Tardigrade Admin · Accepted Answer

XA=(nA/nA+nB) 0.6=((69/46)/(69)46+(x)18 x = 18

Q. $^{'} X ’$ grams of water is mixed in $69$ grams of ethanol. Mole fraction of ethanol in the resultant solution is $0.6$ . What is the value of $‘ X ’$ in grams

54

36

180

18

$X_{A} = \frac{n _{A}}{n _{A} + n _{B}}$
$0.6 = \frac{\frac{69}{46}}{\frac{69}{46} + \frac{x}{18}}$
$x = 18$

Q. ′X’ grams of water is mixed in 69 grams of ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of ‘X’ in grams

54

36

180

18

XA​=nA​+nB​nA​​ 0.6=4669​+18x​4669​​ x=18

Q. $^{'} X ’$ grams of water is mixed in $69$ grams of ethanol. Mole fraction of ethanol in the resultant solution is $0.6$ . What is the value of $‘ X ’$ in grams

$X_{A} = \frac{n _{A}}{n _{A} + n _{B}}$
$0.6 = \frac{\frac{69}{46}}{\frac{69}{46} + \frac{x}{18}}$
$x = 18$