A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vapourized in a 500.0 mL evacuated flask.The pressure of the resulting vapour was 12

Question

A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vapourized in a 500.0 mL evacuated flask.The pressure of the resulting vapour was 12

A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vapourized in a 500.0 mL evacuated flask.The pressure of the resulting vapour was 12.5 mm of Hg at `802^@C`.The structure of the solid metal is known to be body-centered cubic.What is the atomic radius of the metal atom in picometers ?
(R=0.082 It-atm/mol-K)
(The radii of metals atoms as Li=152 pm, Na=186 pm,K=227 pm,Rb=248 pm)

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 231pm
n=no of moles `=(PV)/(RT)=12.5/760xx1/2xx1/(0.082xx1075)=9.318xx10^(-5)`
Number of atoms`=nN_A=5.612xx10^19`
So number of unit cells`=((nN_A)/2)=2.806xx10^19`
So, if there are x unit cells along one edge of given cube.then
Then total number of unit cells in the cubic crystal `=x^3=(nN_A)/2=2.806xx10^19 implies x=3.04xx10^6`
Now if edge length of unit cell =a `implies ax=1.62 mm implies a=5.33xx10^(-9)mm implies a=533` pm
`:. r=(sqrt3a)/4=230.3` pm