Question

1.
The number ofpie electrons present in 2.6 g of ethyne
(1) 2.40 x 1024
(3) 2.40 x 1023
(2) 2.40 * 1025
(4) 1.20 x 1023

Answer 1

$\huge\underline\green{\sf Answer:-}$

$\large{\boxed{\sf No.of Electron = 2.4×{10}^{23}}}$

$\huge\underline\green{\sf Solution:-}$

No. of π- bond = 2

No. of electron involved in 2π bond =4(2 electron each )

No. of moles of ethyne =$\sf{{\frac{given\:mass}{molar\:mass}}}$

$\large{\implies{\sf {\frac{26}{2.6}}}}$

$\large{\sf No.\:of\:mole = 0.1}$

Now ,

Number of electron in 0.1 mole of ethyne

$\large{\implies{\sf N_{A}×moles×no\:of\: Electron}}$

$\large{\implies {\sf 6.023×{10}^{23}×0.1×4}}$

$\large{\boxed{\sf No.\:of\:moles\:in\:0.1\:mole\:ethyne=2.4×{10}^{23}}}$