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प्रश्न
The observed dipole-moment of HCl molecule is 1.03 D. If \[\ce{H - Cl}\] bond distance is 1.275 Å and electronic charge is 4.8 × 10-10 esu. What is the percent polarity of HCl?
विकल्प
1.275 Å × 1.03
`(4.8 xx 10^-10 xx 1.275 xx 10^-8)/1.03`
`(1.03 xx 100 xx 10^-18)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`
`(4.8 xx 10^-10)/1.03 xx 10`
MCQ
रिक्त स्थान भरें
उत्तर
`(1.03 xx 100 xx 10^-18)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`
Explanation:
`"m"_"calculated" = "e" xx "d"`
= `4.8 xx 10^-10 xx 1.275 xx 10^-8`
d = `1.25 xx 10^-8` cm
1 D = `1 xx 10^-18` esu - cm
`mu_"observed" = 1.03` D
= `1.03 xx 10^-18` esu-cm
% = `mu_"observed"/mu_"cal" xx 100`
`= (1.03 xx 10^-18 xx 100)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`
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