The van der Waals’ constant ‘b’ of a glass is 4π×10−4L/mol. The centers of the two molecules approach each other? [Use:NA=6×1023] by 10−n then n is______

b=4×43×πr3×NA;4×π×10−4×1000=4×43×π×r3×6×1023,r=5×10−9 cmDistance of closest approach =2r10−8 cm or 10−10m

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The van der Waals’ constant ‘b’ of a glass is $4 π \times 10^{- 4} L / m o l .$ The centers of the two molecules approach each other? $[U s e : N_{A} = 6 \times 10^{23}]$ by $10^{- n} t h e n n i s______$

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Detailed Solution

$b = 4 \times \frac{4}{3} \times π r^{3} \times N_{A}; 4 \times π \times 10^{- 4} \times 1000 = 4 \times \frac{4}{3} \times π \times r^{3} \times 6 \times 10^{23}, r = 5 \times 10^{- 9} c m$

Distance of closest approach =2r

$10^{- 8} c m o r 10^{- 10} m$

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