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Solutions question

2021 · 25 Jul · Shift 1 · Q18
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Solutions question

2021 · 25 Jul · Shift 1 · Q18

JEE MainChemistrySolutionsNumerical+4 / −1
CO2CO_2CO2​ gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2CO_2CO2​ exerts a partial pressure of 0.835 bar then x m mol of CO2CO_2CO2​ would dissolve in 0.9 L of water. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) (Henry's law constant for CO2CO_2CO2​ at 298 K is 1.67 ×\times× 103 bar)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Use Henry’s law

    For dissolution of a gas in a liquid, p=KHxp = K_H xp=KH​x where:

    • ppp = partial pressure of gas
    • KHK_HKH​ = Henry’s law constant
    • xxx = mole fraction of dissolved gas

  2. Substitute the given values

    p=0.835 bar,KH=1.67×103 barp = 0.835\ \text{bar}, \qquad K_H = 1.67 \times 10^3\ \text{bar}p=0.835 bar,KH​=1.67×103 bar

    So, x=pKH=0.8351.67×103x = \frac{p}{K_H} = \frac{0.835}{1.67 \times 10^3}x=KH​p​=1.67×1030.835​

    x=5.0×10−4x = 5.0 \times 10^{-4}x=5.0×10−4

  3. Relate mole fraction to moles dissolved

    Let moles of dissolved CO2=nCO_2 = nCO2​=n.

    Moles of water in 0.9 L0.9\,L0.9L:

    Since density of water ≈1 g/mL\approx 1\,g/mL≈1g/mL, 0.9 L=900 mL⇒900 g0.9\,L = 900\,mL \Rightarrow 900\,g0.9L=900mL⇒900g

    moles of water=90018=50\text{moles of water} = \frac{900}{18} = 50moles of water=18900​=50

    Mole fraction of CO2CO_2CO2​ is x=nn+50x = \frac{n}{n+50}x=n+50n​

    Since x=5.0×10−4x = 5.0 \times 10^{-4}x=5.0×10−4 is very small, we may take x≈n50x \approx \frac{n}{50}x≈50n​

    Thus, n=50×5.0×10−4=2.5×10−2 moln = 50 \times 5.0 \times 10^{-4} = 2.5 \times 10^{-2}\,\text{mol}n=50×5.0×10−4=2.5×10−2mol

  4. Convert to mmol

    2.5×10−2 mol=25 mmol2.5 \times 10^{-2}\,\text{mol} = 25\,\text{mmol}2.5×10−2mol=25mmol

  5. Nearest integer

    x=25x = 25x=25

Verification with stored answer

Stored correct answer = 252525

Our derived answer = 252525

So, the answer agrees.

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