x grams of water is mixed in 69 g 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 g 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) 18 (D) 9

Tardigrade Admin · Accepted Answer

wA=xg, mA=18, xA=1-0.6=0.4 wB=69g, mB=46, xB=0.6, xA=(nA/nA+nB) 0.4=(wA/mA/wA/mA+w2/mB)=(x/18/(x)18+(69)45 x=18

Q. $x$ grams of water is mixed in 69 g ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of $x$ in grams?

54

36

18

9

$w_{A} = xg,$ $m_{A} = 18,$ $x_{A} = 1 - 0.6 = 0.4$ $w_{B} = 69 g,$ $m_{B} = 46,$ $x_{B} = 0.6,$ $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ $0.4 = \frac{w _{A} / m _{A}}{w _{A} / m _{A} + w _{_{2}} / m _{B}} = \frac{x /18}{\frac{x}{18} + \frac{69}{45}}$ $x = 18$

Q. x grams of water is mixed in 69 g ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of x in grams?

54

36

18

9

wA​=xg, mA​=18, xA​=1−0.6=0.4 wB​=69g, mB​=46, xB​=0.6, xA​=nA​+nB​nA​​ 0.4=wA​/mA​+w2​​/mB​wA​/mA​​=18x​+4569​x/18​ x=18

Q. $x$ grams of water is mixed in 69 g ethanol. Mole fraction of ethanol in the resultant solution is 0.6. What is the value of $x$ in grams?

$w_{A} = xg,$ $m_{A} = 18,$ $x_{A} = 1 - 0.6 = 0.4$ $w_{B} = 69 g,$ $m_{B} = 46,$ $x_{B} = 0.6,$ $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ $0.4 = \frac{w _{A} / m _{A}}{w _{A} / m _{A} + w _{_{2}} / m _{B}} = \frac{x /18}{\frac{x}{18} + \frac{69}{45}}$ $x = 18$