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Question
Total number of electrons present in 1.7 g of ammonia is
Options
6.022 × 1023
`(6.022 xx 10^22)/1.7`
`(6.022 xx 10^24)/1.7`
`(6.022 xx 10^23)/1.7`
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Solution
6.022 × 1023
Explanation:
No. of electrons present in one ammonia (NH3) molecule (7 + 3) = 10
No. of moles of ammonia = `"Mass"/"Molar mass"`
= `(1.7 "g")/(17 "gmol"^-1)`
= 0.1 mol
No. of molecules present in 0.1 mol of ammonia
= 0.1 × 6.022 × 1023
= 6.022 × 1022
No. of electrons present in 0.1 mol of ammonia
= 10 × 6.022 × 1022
= 6.022 × 1023
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