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Current Question (ID: 17932)

Question:

$\text{A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm}^{-3}\text{. The molar mass of the metal is:}$ $\left( N_A \text{ Avogadro's constant} = 6.02 \times 10^{23} \text{ mol}^{-1} \right)$

Options:

1. $30 \text{ g mol}^{-1}$
2. $27 \text{ g mol}^{-1}$
3. $20 \text{ g mol}^{-1}$
4. $40 \text{ g mol}^{-1}$

Solution:

$\text{Hint: } d = \frac{zM}{N_Aa^3}$ $\text{Complete step by step answer:}$ $\text{Step 1: In this step lets enlist all the given quantities: Density, } d=2.72 \text{ g/cm}^3$ $z=\text{number of effective constituent particles in one unit cell}$ $M=\text{molecular weight}$ $a=404 \text{ pm edge length of unit cell}$ $N_A = 6.023 \times 10^{23}$ $\text{Step 2: In this step we will use the formula for density to find } M \text{ (molecular weight):}$ $\text{The formula for density is given by: } d = \frac{zM}{a^3N_A}$ $\Rightarrow M = \frac{da^3N_A}{z}$ $\text{we know that for an fcc structure, } z=4$ $\Rightarrow M = \frac{2.72 \times (404 \times 10^{-10})^3 \times 6.023 \times 10^{23}}{4}$ $\Rightarrow M = 27 \text{ g/mol}$ $\text{So here we got the required molecular mass and with the correct units.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}