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Solutions question

2024 · 29 Jan · Shift 1 · Q30
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Solutions question

2024 · 29 Jan · Shift 1 · Q30

JEE MainChemistrySolutionsNumerical+4 / −1
The osmotic pressure of a dilute solution is 7×105 Pa7 \times 10^5 \mathrm{~Pa}7×105 Pa at 273 K273 \mathrm{~K}273 K. Osmotic pressure of the same solution at 283 K283 \mathrm{~K}283 K is ‾\underline{\hspace{2cm}}​×104Nm−2\times 10^4 \mathrm{Nm}^{-2}×104Nm−2.
Numerical answer
View written solutionFree

Correct answer: 73

  1. Use the relation for osmotic pressure

For a dilute solution,

π=CRT\pi = CRTπ=CRT

where CCC and RRR remain constant for the same solution.

So,

π∝T\pi \propto Tπ∝T
  1. Set up the ratio

Given:

π1=7×105 Pa at T1=273 K\pi_1 = 7 \times 10^5\ \text{Pa at } T_1 = 273\ \text{K}π1​=7×105 Pa at T1​=273 K

At T2=283 KT_2 = 283\ \text{K}T2​=283 K,

π2π1=T2T1\frac{\pi_2}{\pi_1} = \frac{T_2}{T_1}π1​π2​​=T1​T2​​

Therefore,

π2=π1⋅T2T1=7×105⋅283273\pi_2 = \pi_1 \cdot \frac{T_2}{T_1} = 7 \times 10^5 \cdot \frac{283}{273}π2​=π1​⋅T1​T2​​=7×105⋅273283​
  1. Calculate
283273≈1.03663\frac{283}{273} \approx 1.03663273283​≈1.03663

So,

π2≈7×105×1.03663=7.2564×105 Pa\pi_2 \approx 7 \times 10^5 \times 1.03663 = 7.2564 \times 10^5\ \text{Pa}π2​≈7×105×1.03663=7.2564×105 Pa
  1. Convert into the asked form

We need:

π2=‾×104 Nm−2\pi_2 = \underline{\hspace{1cm}} \times 10^4\ \text{Nm}^{-2}π2​=​×104 Nm−2

Since,

7.2564×105=72.564×1047.2564 \times 10^5 = 72.564 \times 10^47.2564×105=72.564×104

For an integer answer, this is approximately

73×104 Nm−273 \times 10^4\ \text{Nm}^{-2}73×104 Nm−2
  1. Final answer
73\boxed{73}73​
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