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)

A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vapo

1.	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)
Answer» SOLUTION :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