Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Vapour pressure of water at 293 K is 17.535 mm Hg.Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

2.34    Vapour pressure of water at  293 \; K is  17.535\; mm\; Hg. Calculate the vapour pressure of water at 293 \; K  when  25 \; g  of glucose is dissolved in 450 \; g of water.

Answers (1)

best_answer

Firstly we will find number of moles of both water and glucose.

Moles of glucose :

                                 = \frac{25}{180} = 0.139\ mol                                        (Molar\ mass\ of\ glucose = 6(12)+12(1)+6(16) = 180\ g\ mol^{-1})

and moles of water :

                                   = \frac{450}{18} = 25\ mol                                            (Molar\ mass\ of\ water = 18\ g\ mol^{-1})

Now,

                                             \frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_g}{n_g + n_w}

or                                        \frac{17.535 - p}{17.535} = \frac{0.139}{0.139+ 25}

or                                          p = 17.44\ mm\ of\ Hg 

Thus vapour pressure of water after glucose addition = 17.44 mm of Hg

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date Sheet 2025: CBSE, UPMSP, MPBSE, Class 10, 12 board exam dates out; BSEB, CGBSE schedules soon
anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

Latest Question