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X-ray diffraction studies show that edge length of a unit cell of NaCl is 0.56 nm. Density of NaCl was found to be `2.16g//"cc".` What type of defect is found in the solid? Calculate the percentage of `Na^(+)` and `Cl^(-)` ions that are missing. |
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Answer» Correct Answer - Schottky defect; `2.375%` each Density of NaCl `(rho)=(ZxxM)/(a^(3)xxN_(0))=(4xx(58.5"g mol"^(-1)))/((0.56xx10^(-7)cm)^(3)xx(6.022xx10^(23)mol^(-1)))=2.212g cm^(-3)` The observed density `(2.16"g cm"^(-3))` is less than theoretical density. Therefore, the Solid has schottky defect. `Z=(a^(3)xxrhoxxNo)/(M)=((0.56xx10^(-7)cm)^(3)xx(2.16"g cm"^(-3))xx(6.023xx10^(23)"g mol"^(-1)))/((58.5"g mol"^(-1)))=3.905` Number of missing formula units `=4-3.905=0.095` Percentage of missing formula units `=(0.095)/(4)xx100=2.375%` Percentage of missing `Na^(+)` ions `=2.375%` Percentage of missing `K^(+)` ions `=2.375%` |
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