Import Question JSON

Current Question (ID: 17950)

Question:

$\text{Niobium crystallizes in body-centred cubic structure. If the density is } 8.55 \text{ g cm}^{-3}, \text{ the atomic radius of niobium (atomic mass 93 u) is -}$

Options:

1. $13.32 \text{ nm}$
2. $12.32 \text{ nm}$
3. $14.32 \text{ nm}$
4. $15.32 \text{ nm}$

Solution:

$\text{STEP 1: The given data is as follows:}$ $\text{It is given that the density of niobium, } d = 8.55 \text{ g cm}^{-3}$ $\text{Atomic mass, } M = 93 \text{ g mol}^{-1}$ $\text{As the lattice is bcc type, the number of atoms per unit cell, } z = 2$ $\text{We know that } N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$ $\text{STEP 2: Calculate edge length using density formula.}$ $\text{Applying the relation:}$ $d = \frac{zM}{a^3N_A}$ $\Rightarrow a^3 = \frac{zM}{dN_A}$ $= \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} \text{ g mol}^{-1} \times \text{ g cm}^{-3} \times \text{ mol}^{-1}$ $= 3.612 \times 10^{-23} \text{ cm}^3$ $\text{So, } a = 3.306 \times 10^{-8} \text{ cm}$ $\text{STEP 3: Calculate the atomic radius using edge length value as follows:}$ $\text{For body-centered cubic unit cell:}$ $r = \frac{\sqrt{3}}{4}a$ $= \frac{\sqrt{3}}{4} \times 3.306 \times 10^{-8} \text{ cm}$ $= 1.432 \times 10^{-8} \text{ cm}$ $\text{as, } 1 \text{ m} = 100 \text{ cm}$ $r = 1.432 \times 10^{-8} \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}}$ $= 1.432 \times 10^{-10} \text{ m}$ $= 14.32 \times 10^{-9} \text{ m}$ $r = 14.32 \times 10^{-9} \text{ m}$ $\text{as, } 1 \text{ m} = 10^{9} \text{ nm}$ $r = 14.32 \text{ nm}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}