Question

Chemistry (Class-11 & Chapter-1) Q. How many electrons are present in 42g ${N_3}^{ - }$ ?
(a) 21 × 6.022 × 10^23
(b) 22 × 6.022 × 10^23
(c) 15 × 6.022 × 10^23
(d) 16 × 6.022 × 10^23​

Answer 1

=9.6352* 10^23 valence electrons.

ANSWER:-

Given mass= 4.2g

Molar mass of N3^- (azide ion) = 14*3= 42

Therefore, No. Of moles of azide ion = given mass/ molecular mass= 4.2/42= 0.1 moles

No. of valence electrons in one nitrogen atom = 5

No. of valence electrons in azide ion (N3^-) ( N=N=N ) is =(5*3)+1=16electrons

Therefore, total no. of valence electrons in 4.2g of azide (N3^-) ion = no. of moles of azide ion * no. of valence electrons in one azide ion * Avogadro's constant

= 0.1* 16* 6.022*10^23

=1.6* 6.022*10^ 23

=9.6352* 10^23 valence electrons.

Hope this helps you

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Answer 2

Answer:

REFER THE ATTACHMENT

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