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The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is: (Given: Molar Mass Na: 23 and Cl:…

Solutions · Class 12 · JEE Main Previous Year Question

Question

The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is:

(Given: Molar Mass Na: 23 and Cl: 35.5 g mol⁻¹)

Options
  1. a

    2

  2. b

    20

  3. c

    4

  4. d

    0.2

Correct Answerd

0.2

Detailed Solution

Strategy:\n> Molarity (MM) is a concentration term that defines the number of moles of solute present in exactly one liter of a solution. Calculate the moles of \ceNaCl\ce{NaCl} and divide by the volume in liters.\n\nStep 1: Calculate the molar mass of } \ce{NaCl} \text{\n- Atomic mass of \ceNa=23textg/mol\ce{Na} = 23\\text{ g/mol}\n- Atomic mass of \ceCl=35.5textg/mol\ce{Cl} = 35.5\\text{ g/mol}\n- Molar mass =23+35.5=58.5textg/mol= 23 + 35.5 = 58.5\\text{ g/mol}\n\nStep 2: Determine moles of } \ce{NaCl} \text{\nn=fracwM=frac5.85textg58.5textg/mol=0.1textmoln = \\frac{w}{M} = \\frac{5.85\\text{ g}}{58.5\\text{ g/mol}} = 0.1\\text{ mol}\n\nStep 3: Calculate Molarity (MM)\nVolume = 500 mL = 0.5 L.\nM=frac0.1textmol0.5textL=0.2textMM = \\frac{0.1\\text{ mol}}{0.5\\text{ L}} = 0.2\\text{ M}\n\ntextAnswer:(4)\boxed{\\text{Answer: (4)}}

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The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is: (Given: Molar… (JEE Main 2024) | Canvas Classes