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Current Question (ID: 8153)

Question:

Two flasks of equal volume are connected by a narrow tube (of negligible volume) at $27^\circ\text{C}$ and contain $0.70$ mole of $\text{H}_2$ at $0.5$ atm. One of the flasks is then immersed in a hot bath, kept at $127^\circ\text{C}$, while the other remains at $27^\circ\text{C}$. The final pressure is -

Options:

1. $5.714 \, \text{atm}$
2. $0.5714 \, \text{atm}$
3. $2.5214 \, \text{atm}$
4. $5.5114 \, \text{atm}$

Solution:

**Hint:** Use $PV = nRT$ **Step 1:** Each flask initially contains $0.35$ mole of $\text{H}_2$. Let ‘$x$’ moles of hydrogen gas diffuse from flask II (hot) to flask I (cold). Final moles in flask I: $(0.35 + x)$ Final moles in flask II: $(0.35 - x)$ Using the ideal gas law for both flasks at equilibrium: For flask I (cold, $T_1 = 300\, \text{K}$): $PV = (0.35 + x) \times R \times 300$ For flask II (hot, $T_2 = 400\, \text{K}$): $PV = (0.35 - x) \times R \times 400$ Equating the two expressions (since $P$ and $V$ are the same): $(0.35 + x) \times 300 = (0.35 - x) \times 400$ Solving gives $x = 0.05$ moles. **Step 2:** Total moles after diffusion: $(0.35 + 0.05) + (0.35 - 0.05) = 0.70$ (unchanged) Using the ratio of pressures before and after: $\frac{P_1}{P_2} = \frac{n_1 T_1}{n_2 T_2}$ Here, $n_1 = 0.35$, $T_1 = 300\, \text{K}$, $n_2 = 0.40$, $T_2 = 300\, \text{K}$ (for flask I) But a simpler approach is to calculate the new pressure directly: $P = \frac{n_{\text{total}} R T_{\text{avg}}}{V_{\text{total}}}$ However, from the given solution: $\frac{0.5}{P} = \frac{0.35}{0.40}$ $P = 0.5714 \, \text{atm}$

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{
  "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
}