A) 34.84%
B) 43.8%
C) 23.4%
D) 21.6%
Correct Answer: A
Solution :
Let the volume of solid \[\text{Ar }=\text{ 1}00\text{ mL}\] Mass of solid \[Ar=V\times \rho =100\times 1.59g/c{{m}^{3}}\] \[=159g\] Volume of liquid Ar \[=\frac{Mass}{\rho }=\frac{159g}{1.4gc{{m}^{-3}}}\] \[=11.357\text{ml}\] \[\because \] Ar crystallizes in fcc type lattice, the packing fraction \[=0.74\] Actual volume occupied by Ar \[\text{= Packing fraction }\!\!\times\!\!\text{ Volume of solid Ar}\]\[\text{= 0}\text{.74 }\!\!\times\!\!\text{ 100 =74 mL}\] \[\therefore \]% empty space in liquid \[Ar=\frac{(11357-74)\times 100}{113.57}\] \[=34.84%\]You need to login to perform this action.
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