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Current Question (ID: 8234)

Question:

$\text{The temperature at which the r.m.s velocity of carbon dioxide becomes the same as that of nitrogen at }21^0\text{C is-}$

Options:

1. $462 ^{\circ}\text{C}$
2. $273 \text{ K}$
3. $189 ^{\circ}\text{C}$
4. $546 \text{ K}$

Solution:

$\text{Hint: }v_{\text{rms}} = \sqrt{\frac{3\text{RT}}{\text{M(molar mass)}}}$ $\text{Step 1:}$ $\text{The formula of }v_{\text{rms}}\text{ is as follows:}$ $v_{\text{rms}} = \sqrt{\frac{3\text{RT}}{\text{M(molar mass)}}}$ $\text{The }v_{\text{rms}}\text{ of carbon dioxide is equal to }v_{\text{rms}}\text{ of nitrogen.}$ $\text{Step 2:}$ $\text{Calculate the temperature at which the r.m.s velocity of carbon dioxide becomes the same as that of nitrogen as follows:}$ $v_{\text{rms}}\text{ of carbon dioxide = }v_{\text{rms}}\text{ of nitrogen.}$ $\frac{\sqrt{3\text{RT}_1}}{\text{M}_1} = \frac{\sqrt{3\text{RT}_2}}{\text{M}_2}$ $\frac{\text{T}_1}{\text{M}_1} = \frac{\text{T}_2}{\text{M}_2}$ $\text{T}_1 = \text{T}_2 \times \frac{\text{M}_1}{\text{M}_2}$ $\text{T}_1 = 294 \times \frac{44}{28} = 462 \text{ K}$ $= 189 ^{\circ}\text{C}$

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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
}