100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH.The volume of Ethane obtained at STP after the complete electrolysis of the resulting solution is

Question

100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH.The volume of Ethane obtained at STP after the complete electrolysis of the resulting solution is (A) 112 mL (B) 56 mL (C) 224 mL (D) 560 mL

Tardigrade Admin · Accepted Answer

Acetic acid on electrolysis gives ethane as

Given moles of acetic acid,

C H_{3} COO H = \frac{0.1 \times 100}{1000} = 0.01

∵ 2 m o lC H_{3} COO H

gives

C_{2} H_{6} = 22.4 L

∴ 0.01 m o lC H_{3} COO H

will give

C_{2} H_{6} = \frac{22.4 \times 0.01}{2}

= 0.112 L

= 112 m L

Q. 100mL of 0.1M acetic acid is completely neutralized using a standard solution of NaOH.The volume of Ethane obtained at STP after the complete electrolysis of the resulting solution is

112 mL

56 mL

224 mL

560 mL

Acetic acid on electrolysis gives ethane as Given moles of acetic acid, CH3​COOH=10000.1×100​=0.01 ∵2molCH3​COOH gives C2​H6​=22.4L ∴0.01molCH3​COOH will give C2​H6​=222.4×0.01​ =0.112L =112mL

Q. $100 m L$ of $0.1 M$ acetic acid is completely neutralized using a standard solution of $N a O H$ .The volume of Ethane obtained at $STP$ after the complete electrolysis of the resulting solution is

Acetic acid on electrolysis gives ethane as

Given moles of acetic acid,

$C H_{3} COO H = \frac{0.1 \times 100}{1000} = 0.01$

$∵ 2 m o lC H_{3} COO H$ gives $C_{2} H_{6} = 22.4 L$

$∴ 0.01 m o lC H_{3} COO H$ will give

$C_{2} H_{6} = \frac{22.4 \times 0.01}{2}$

$= 0.112 L$

$= 112 m L$