14.96 g of a mixture containing n-hexane and ethyl alcohol are reacted with sodium metal to give 200 ml of H 2 at 27° C and 760 mm pressure. The percentage of ethyl alcohol in the mixture is . [ R =0.082 L . atm mol -1 K -1]

Question

14.96 g of a mixture containing n-hexane and ethyl alcohol are reacted with sodium metal to give 200 ml of H 2 at 27° C and 760 mm pressure. The percentage of ethyl alcohol in the mixture is . [ R =0.082 L . atm mol -1 K -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Only ethyl alcohol reacts with sodium metal.

C_{2} H_{5} O H + N a \to C_{2} H_{5} ON a + 1/2 H_{2}

No. of moles of

H_{2}

formed

= \frac{P V}{RT}

= \frac{760}{760} \times \frac{200}{1000} \times \frac{1}{0.082 \times 300}

= \frac{1}{15 \times 8.2} - \frac{1}{123} - 8.13 \times 1 0^{- 3}

0.5

mole of

H_{2}

is liberated by

1

mole of

C_{2} H_{5} O H

.

8.13 \times 1 0^{- 3}

mole

H_{2}

is liberated by

\frac{1 \times 8.13 \times 1 0 ^{- 3}}{0.5}

mole of

C_{2} H_{5} O H

= 16.26 \times 1 0^{- 3}

mole of

C_{2} H_{3} O H

= 16.26 \times 1 0^{- 3} \times 46 g

of

C_{2} H_{5} O H

= 0.748 g

The percentage of ethyl alcohol

n

the mixture

= \frac{7.48 \times 100}{14.96} = \frac{74.8}{14.96} = 5%

Q. 14.96g of a mixture containing n-hexane and ethyl alcohol are reacted with sodium metal to give 200ml of H2​ at 27∘C and 760mm pressure. The percentage of ethyl alcohol in the mixture is ______. [R=0.082L atmmol−1K−1]

Q. $14.96 g$ of a mixture containing $n$ -hexane and ethyl alcohol are reacted with sodium metal to give $200 m l$ of $H_{2}$ at $2 7^{\circ} C$ and $760 mm$ pressure. The percentage of ethyl alcohol in the mixture is ______. $[R = 0.082 L$ $a t m m o l^{- 1} K^{- 1}]$