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Current Question (ID: 18044)

Question:

$\text{The density of 68\% nitric acid by mass in an aqueous solution is 1.504 g mL}^{-1}. \text{ The molarity of the acid solution would be:}$

Options:

1. $15.24 \text{ M}$
2. $16.23 \text{ M}$
3. $14.52 \text{ M}$
4. $13.45 \text{ M}$

Solution:

$\text{Hint: molarity} = \frac{\text{amount}}{\text{molar mass of nitric acid}} \times \frac{1000}{\text{volume of solution(mL)}}$ $\text{Step 1:}$ $\text{Find the volume of the solution.}$ $\text{The amount of nitric acid is 68\% w/w. It indicates that 68 g nitric acid presents at 100 g of solution.}$ $\text{The amount of solution is 100g.}$ $\text{Calculate the volume of solution using the density formula is:}$ $\text{density} = \frac{\text{amount of solution}}{\text{volume of solution}}$ $1.504 \text{ g mL}^{-1} = \frac{100 \text{ g}}{\text{volume of solution}}$ $\text{volume of solution} = \frac{100}{1.504}$ $\text{volume of solution} = 66.5 \text{ mL}$ $\text{Step 2:}$ $\text{Calculate the molarity of the solution is as follows:}$ $\text{molarity} = \frac{\text{amount of nitric acid}}{\text{molar mass of nitric acid}} \times \frac{1000}{\text{volume of solution(mL)}}$ $\text{molarity} = \frac{68}{63} \times \frac{1000}{66.5}$ $\text{molarity} = 16.23 \text{ M}$

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
}