JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Solid State

KBr is doped with

1 0^{- 5}

mole percent of

SrBr_{2}

. The number of cationic vacancies in 1 g of KBr crystal is ______

\times 1 0^{14}

. (Round off to the nearest integer).
[Atomic Mass: K:

39.1 u

, Br:

79.9 u

N_{A} = 6.023 \times 1 0^{23}

]

Answer: 5

Solution

Potassium bromide has molar mass

39.1 + 79.9 = 119 g mo l^{- 1}

, so one gram of the crystal is

n = \frac{1}{119} = 8.403 \times 1 0^{- 3} mol

A doping level of

1 0^{- 5}

mole percent means

1 0^{- 5}

mol of

SrB r_{2}

for every

100

mol of the host, so the amount of the dopant present is

8.403 \times 1 0^{- 3} \times \frac{1 0 ^{- 5}}{100} = 8.403 \times 1 0^{- 10} mol

Each

S r^{2 +}

ion carries the charge of two

K^{+}

ions but occupies only one cation site, so one cation vacancy appears for every strontium ion introduced. The number of vacancies is

8.403 \times 1 0^{- 10} \times 6.023 \times 1 0^{23} = 5.06 \times 1 0^{14}

so there are

5 \times 1 0^{14}

cationic vacancies.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer