Import Question JSON

Current Question (ID: 8145)

Question:

$ \text{The volume of } \text{H}_2 \text{ that would be released when 0.15 g of aluminium reacts at 20 } ^\circ \text{C and 1 bar pressure with caustic soda is-} $

Options:

1. $ \text{2.25 L} $
2. $ \text{301 mL} $
3. $ \text{1.56 L} $
4. $ \text{203 mL} $

Solution:

$ \text{Hint: At STP volume of one mole of any gas is 22.4L.} \n\n \text{Step 1: Write down the balanced equation and find out the volume of } \text{H}_2 \text{ at STP.} \n\n \text{The reaction of aluminium with caustic soda can be represented as:} \n\n 2 \text{Al} + 2 \text{NaOH} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaAlO}_2 + 3 \text{H}_2 \n 2 \times 27 \text{g} \quad \quad \quad \quad \quad \quad 3 \times 22400 \text{ mL} \n\n \text{At STP (273.15K and 1 atm), 54g (2 } \times \text{ 27g) of Al gives 3 } \times \text{ 22400 mL of } \text{H}_2 \n \Rightarrow \text{0.15 g Al gives } \frac{3 \times 22400 \times 0.15}{54} \text{ of } \text{H}_2 \text{ i.e., 186.67 mL of } \text{H}_2 \n\n \text{Step 2: Use relation } \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \text{ to find the required volume.} \n\n \text{At STP;} P_1 = 1 \text{ atm, } V_1 = 186.67 \text{ mL, } T_1 = 273.15 \text{ K} \n\n \text{Let the volume of dihydrogen be } V_2 \text{ at } P_2 = 0.987 \text{ atm (since 1 bar = 0.987 atm) and T} = 20 ^\circ \text{C} = (273.15 + 20) = 293.15 \text{ K.} \n\n \text{Now,} \n\n \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \n\n V_2 = \frac{P_1V_1T_2}{P_2T_1} \n\n = \frac{1 \times 186.67 \times 293.15}{0.987 \times 273.15} \n\n V_2 = 202.98 \text{ mL} \approx 203 \text{ mL} $

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
}