Question
An alkali metal has density 4.5 g/cm3. It has cubic unit cell with edge length 400 pm. The reaction of 7.68 cm3 chunk of the metal with an excess of HCl solution gives a colourless gas which occupies 4.54 L at 0°C and 1 bar. The unit cell of metal is
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Solution
$M + HCl \rightarrow MCl + \frac{1}{2}H_{2}$
1 mole (1/2) mole
= A gm $= \frac{1}{2} \times 22.7\text{L at }\text{0}^{0}C\text{ and 1 bar}$
$\therefore(7.68 \times 4.5)\text{ gm}\frac{11.35}{A} \times (7.68 \times 4.5) = 4.54 $$${\therefore A = 86.4 }{\text{Now},d = \frac{Z.M}{N_{A}.V} \Rightarrow 4.5 = \frac{Z \times 86.4}{\left( 6 \times 10^{23} \right) \times \left( 400 \times 10^{- 10} \right)^{3}} }{\therefore Z = 2 }{\Rightarrow \text{ Unit cell is BCC.}}$$
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