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Solutions question

2023 · 11 Apr · Shift 2 · Q8
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Solutions question

2023 · 11 Apr · Shift 2 · Q8

JEE MainChemistrySolutionsMCQ+4 / −1
What weight of glucose must be dissolved in 100 g100 \mathrm{~g}100 g of water to lower the vapour pressure by 0.20 mm Hg0.20 \mathrm{~mm} ~\mathrm{Hg}0.20 mm Hg? (Assume dilute solution is being formed) Given : Vapour pressure of pure water is 54.2 mm Hg54.2 \mathrm{~mm} ~\mathrm{Hg}54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol−1180 \mathrm{~g} \mathrm{~mol}^{-1}180 g mol−1
  1. A
    3.69 g
  2. B
    2.59 g
  3. C
    3.59 g
  4. D
    4.69 g
View written solutionFree

Correct answer: A

  1. Use relative lowering of vapour pressure for a non-volatile solute:

p0−pp0=xsolute\frac{p^0-p}{p^0}=x_{\text{solute}}p0p0−p​=xsolute​

Since the solution is dilute,

xsolute≈n2n1x_{\text{solute}}\approx \frac{n_2}{n_1}xsolute​≈n1​n2​​

where:

  • p0=54.2 mm Hgp^0 = 54.2\,\text{mm Hg}p0=54.2mm Hg
  • p0−p=0.20 mm Hgp^0-p = 0.20\,\text{mm Hg}p0−p=0.20mm Hg
  • n1n_1n1​ = moles of water
  • n2n_2n2​ = moles of glucose
  1. Calculate relative lowering of vapour pressure:

Δpp0=0.2054.2=0.00369\frac{\Delta p}{p^0}=\frac{0.20}{54.2}=0.00369p0Δp​=54.20.20​=0.00369

So,

xsolute≈n2n1=0.00369x_{\text{solute}} \approx \frac{n_2}{n_1}=0.00369xsolute​≈n1​n2​​=0.00369

  1. Calculate moles of water:

Mass of water =100 g=100\,\text{g}=100g

n1=10018=5.56 moln_1=\frac{100}{18}=5.56\,\text{mol}n1​=18100​=5.56mol

  1. Find moles of glucose:

n2=0.00369×5.56=0.0205 moln_2 = 0.00369 \times 5.56 = 0.0205\,\text{mol}n2​=0.00369×5.56=0.0205mol

  1. Convert moles of glucose to mass:

Molar mass of glucose =180 g mol−1=180\,\text{g mol}^{-1}=180g mol−1

m=n2×180=0.0205×180=3.69 gm = n_2 \times 180 = 0.0205 \times 180 = 3.69\,\text{g}m=n2​×180=0.0205×180=3.69g

  1. Check options:
  • A: 3.69 g3.69\,\text{g}3.69g ✅
  • B: 2.59 g2.59\,\text{g}2.59g ❌
  • C: 3.59 g3.59\,\text{g}3.59g ❌
  • D: 4.69 g4.69\,\text{g}4.69g ❌

Therefore, the correct answer is Option A.

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