KBr crystallises in NaCl type of crystal lattice and its density is 2.75 g / cm 3. Number of unit cells of KBr present in a 1.00 mm 3 grain of KBr are

Question

KBr crystallises in NaCl type of crystal lattice and its density is 2.75 g / cm 3. Number of unit cells of KBr present in a 1.00 mm 3 grain of KBr are (A) 3.6 × 1018 (B) 3.6 × 1015 (C) 2.34 × 1023 (D) 6.02 × 1024

Tardigrade Admin · Accepted Answer

In a unit cell of KBr, there are four formula units of

K B r

therefore,
Density

(ρ) = \frac{4 \times 119}{6.023 \times 1 0 ^{23} \times a ^{3}} = 2.75

a^{3} = 2.873 \times 1 0^{- 22} c m^{3}

= 2.778 \times 1 0^{- 19} m m^{3}

Number of unit cells per

m m^{3}

= \frac{1 0 ^{19}}{2.873} = 3.5 \times 1 0^{18}

Q. $K B r$ crystallises in $N a Cl$ type of crystal lattice and its density is $2.75 g / c m^{3}$ . Number of unit cells of $K B r$ present in a $1.00 m m^{3}$ grain of $K B r$ are

$3.6 \times 1 0^{18}$

$3.6 \times 1 0^{15}$

$2.34 \times 1 0^{23}$

$6.02 \times 1 0^{24}$

Q. KBr crystallises in NaCl type of crystal lattice and its density is 2.75g/cm3. Number of unit cells of KBr present in a 1.00mm3 grain of KBr are

3.6×1018

3.6×1015

2.34×1023

6.02×1024

In a unit cell of KBr, there are four formula units of KBr therefore, Density (ρ)=6.023×1023×a34×119​=2.75 a3=2.873×10−22cm3 =2.778×10−19mm3 Number of unit cells per mm3 =2.8731019​=3.5×1018

Q. $K B r$ crystallises in $N a Cl$ type of crystal lattice and its density is $2.75 g / c m^{3}$ . Number of unit cells of $K B r$ present in a $1.00 m m^{3}$ grain of $K B r$ are

$3.6 \times 1 0^{18}$

$3.6 \times 1 0^{15}$

$2.34 \times 1 0^{23}$

$6.02 \times 1 0^{24}$