Molybdenum (At. mass = 96g mol-l) crystallizes as bcc crystal. If density of crystal is 10.3 g/cm3, then radius of Mo atom is (use NA= 6 x 1023mol-1) .

d=zMWNAa3Z=2,Mw=96,d=10.3g/cm33a=4r=3a4a3=2×9610.3×6.023×1023=3.14×10−8r=3a4=3×3.14×10−84=135.96pm

Molybdenum (At. mass = 96g mol-l) crystallizes as bcc crystal. If density of crystal is 10.3 g/cm3, then radius of Mo atom is (use NA= 6 x 1023mol-1) .

d=zMWNAa3Z=2,Mw=96,d=10.3g/cm33a=4r=3a4a3=2×9610.3×6.023×1023=3.14×10−8r=3a4=3×3.14×10−84=135.96pm

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Molybdenum (At. mass = 96g mol^-l) crystallizes as bcc crystal. If density of crystal is 10.3 g/cm³, then radius of Mo atom is (use N_A= 6 x 10²³mol^-1) .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

111 pm

none of these

314 pm

135.96 pm

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} d = \frac{{zM}_{W}}{N_{A} a^{3}} \\ Z = 2, M_{w} = 96, d = 10 .3 g / {cm}^{3} \\ \sqrt{3} a = 4 r = \frac{\sqrt{3} a}{4} \\ a^{3} = \frac{2 \times 96}{10 .3 \times 6 .023 \times 10^{23}} = 3 .14 \times 10^{- 8} \\ r = \frac{\sqrt{3} a}{4} = \frac{\sqrt{3} \times 3 .14 \times 10^{- 8}}{4} = 135 .96 pm \end{array}$