Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • I need help with The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be

The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be

  • Option 1)

    18.0

  • Option 2)

    342

  • Option 3)

    60

  • Option 4)

    180

 

Answers (1)

As we learned 

 

Relative lowering of vapour pressure -

Relative lowering of vapour pressure on adding a non-volatile solute in a colligative properties.

-

 

 \frac{P^{o}-P_{s}}{P^{0}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\: or\; 0.00713=\frac{71.5/m}{\frac{71.5}{m}+\frac{1000}{18}}

m=180

 


Option 1)

18.0

Option 2)

342

Option 3)

60

Option 4)

180

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Answer Key 2026: NTA reviews April 5 chemistry paper, asks candidates not to pay challenge fee
anam-khan

JEE Mains 2026: NTA removes answer key post on X; over 11 lakh await response sheet
anam-khan

JEE Mains 2026 April: Marks vs percentile trends; predict rank for BTech admissions at NITs, IIITs

Latest Question