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Some Basic Concepts of Chemistry question

2021 · 24 Feb · Shift 1 · Q19
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Some Basic Concepts of Chemistry question

2021 · 24 Feb · Shift 1 · Q19

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x ×\times× 10-1. The value of x is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer)
Numerical answer
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Correct answer: 2

  1. Given data

    • Mass of compound A=4.5 gA = 4.5\,\text{g}A=4.5g
    • Molar mass (MW) of A=90 g mol−1A = 90\,\text{g mol}^{-1}A=90g mol−1
    • Volume of solution =250 mL=0.250 L= 250\,\text{mL} = 0.250\,\text{L}=250mL=0.250L

  2. Calculate number of moles n=massmolar mass=4.590=0.05 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{90} = 0.05\,\text{mol}n=molar massmass​=904.5​=0.05mol

  3. Calculate molarity Molarity is M=nV=0.050.250=0.20 MM = \frac{n}{V} = \frac{0.05}{0.250} = 0.20\,\text{M}M=Vn​=0.2500.05​=0.20M

  4. Match with the given form The molarity is written as x×10−1x \times 10^{-1}x×10−1 Since 0.20=2.0×10−10.20 = 2.0 \times 10^{-1}0.20=2.0×10−1 we get x=2x = 2x=2

  5. Rounded to nearest integer x=2x = 2x=2

Final Answer

2\boxed{2}2​

The derived answer matches the stored correct answer.

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