The density of solid argon is 1.6 gm/ml at -233°C. If the atomic volume of argon is assumed to be 53×10−28, then what % of solid argon is apparently empty space? [Express your answer by dividing with 15]

Volume of solid argon =391.6=24.375/moleVolume of 1 mole atoms of argon =53×10−23×6.023×1023=10.038% vacant =24.375−10.03824.375×100=60% Ans :6015=4

Book Online Demo

Check Your IQ

Try Test

Courses

The density of solid argon is 1.6 gm/ml at $- 233 ° C$ . If the atomic volume of argon is assumed to be $\frac{5}{3} \times 10^{- 28}$ , then what % of solid argon is apparently empty space? [Express your answer by dividing with 15]

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Volume of solid argon $= \frac{39}{1.6} = 24.375 / mole$

Volume of 1 mole atoms of argon

$= \frac{5}{3} \times 10^{- 23} \times 6.023 \times 10^{23} = 10.038$

$\begin{array}{l} % vacant = \frac{24.375 - 10.038}{24.375} \times 100 \\ = 60 % Ans : \frac{60}{15} = 4 \end{array}$

Watch 3-min video & get full concept clarity