Import Question JSON

Current Question (ID: 7343)

Question:

$\text{The number of significant figures present in the answer of the following calculations [(i), (ii), (iii)] are respectively -}$ $\text{1. } 0.02856 \times 298.15 \times 0.112 / 5785$ $\text{2. } 5 \times 5.364$ $\text{3. } 0.0125 + 0.7864 + 0.0215$

Options:

1. $4, 4, 3$
2. $3, 3, 4$
3. $4, 3, 4$
4. $3, 4, 4$

Solution:

$\text{HINT: Use the basic rules to calculate a number of significant figures.}$ $\text{(i) } 0.02856 \times 298.15 \times 0.112 / 5785$ $\text{The least precise number of calculations = 0.112}$ $\text{Number of significant figures in the answer = number of significant figures in the least precise number = 3}$ $\text{(ii) } 5 \times 5.364$ $\text{In the multiplication and division of numbers, the result must be reported with significant figures which are no more than in the number with least significant figures. 5 is a constant and it has a infinite number of significant figure, and 5.364 has four significant figure. Hence, final answer will have four significant figure.}$ $= 5 \times 5.364 = 26.82$ $\text{The final answer will have 4 significant figure}$ $\text{(iii) } 0.0125 + 0.7864 + 0.0215$ $\text{Since the least number of decimal places in each term is four. The answer must contains four digit after decimal.}$ $0.0125 + 0.7864 + 0.0215 = 0.8204$ $\text{The final answer must contains 4 significant figure.}$ $\text{NB= significant number rules of multiplication and division is not applicable on addition/subtraction}$

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
}