224mL of SO2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ( P ( H 2 O ) o = 2 4 m m o f H g ) i s x × 1 0 2 m m of Hg, the value of x is-----------. (Integer answer)

0 7 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    Moles of SO2 224×10322.4

    =0.01 mole

    Moles of NaOH = 0.1 × 0.1

    = 0.01 mole

    SO2+NaOHNaHSO3

    0.01 mole0.01 mole-

    -0.01 mole

    Non-volatile solute is NaHSO3

    Moles of water = 3618=2

    Using ; relative lowering in V.P

    PoPPo=ixB

    Where; ΔP=P0P is lowering in V.P

    ΔP=P0ixB

    i for NaHSO3 = 2

    here; xB=nBnA since solution is dilute

    ΔP=24×2×0.012=0.24

    ΔP=24×102mmHg

    So; x = 24

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post