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Current Question (ID: 9115)

Question:

$\text{The correct relation about the density of Na and K is}$

Options:

1. $\text{Na} > \text{K}$
2. $\text{Na} = \text{K}$
3. $\text{K} > \text{Na}$
4. $\text{Li} > \text{K}$

Solution:

$\text{HINT: Na is more dense than K}$ $\text{Explanation:}$ $\text{The densities of elements of group 1 are quite low.}$ $(\text{Li} = 0.534 \text{ g/cc, Na} = 0.972, \text{K} = 0.559, \text{Rb} = 1.53, \text{Cs} = 1.903). \text{It increases gradually in moving from Li to Cs.}$ $\text{However potassium is lighter than sodium probably due to an unusual increase in atomic size of potassium.}$ $\text{In the case of sodium and potassium the increase in shell size outweighs the pull of the core on the outer shell electron and so potassium is less dense than sodium.}$ $\text{When we move from Na to K, effect of increase in volume is more pronounced as compared to effect of increase in atomic mass. Due to this potassium is less dense than sodium.}$

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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
}