Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25. Hence percentage of ethyl alcohol by weight is

Question

Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25. Hence percentage of ethyl alcohol by weight is (A) 54% (B) 25% (C) 75% (D) 46%

Tardigrade Admin · Accepted Answer

Mole fraction of H2O = 1 - 0.25 = 0.75 (xC2H5OH/XC2H5OH + XH2O) = (nC2H5OH/nC2H5OH + nH2O) or wt. % = (0.25 × 46/0.25 × 46 + 0.75 × 18)× 100 = 46 %.

Q. Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is $0.25$ . Hence percentage of ethyl alcohol by weight is

54%

25%

75%

46%

Mole fraction of $H_{2} O = 1 - 0.25 = 0.75$
$\frac{x _{C_{2} H_{5} O H}}{X _{C_{2} H_{5} O H} + X _{H_{2} O}} = \frac{n _{_{C_{2} H_{5} O H}}}{n _{C_{2} H_{5} O H} + n _{H_{2} O}}$
or wt. $% = \frac{0.25 \times 46}{0.25 \times 46 + 0.75 \times 18} \times 100 = 46%$ .

Q. Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25. Hence percentage of ethyl alcohol by weight is

54%

25%

75%

46%

Mole fraction of H2​O=1−0.25=0.75 XC2​H5​OH​+XH2​O​xC2​H5​OH​​=nC2​H5​OH​+nH2​O​nC2​H5​OH​​​ or wt. %=0.25×46+0.75×180.25×46​×100=46%.

Q. Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is $0.25$ . Hence percentage of ethyl alcohol by weight is

Mole fraction of $H_{2} O = 1 - 0.25 = 0.75$
$\frac{x _{C_{2} H_{5} O H}}{X _{C_{2} H_{5} O H} + X _{H_{2} O}} = \frac{n _{_{C_{2} H_{5} O H}}}{n _{C_{2} H_{5} O H} + n _{H_{2} O}}$
or wt. $% = \frac{0.25 \times 46}{0.25 \times 46 + 0.75 \times 18} \times 100 = 46%$ .