Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, When a liquid that is immiscible with water was steam distilled at 952ºC at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at

When a liquid that is immiscible with water was steam distilled at 952ºC at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2ºC, what is the molar mass of liquid?

  • Option 1)

    7.975 g/mol

  • Option 2)

    166 g/mol

  • Option 3)

    145.8 g/mol

  • Option 4)

    None \: of \: these

 

Answers (1)

best_answer

As we learned 

 

Vapour Pressure of immiscible liquids -

P_{T}= P_{A}^{0}+P_{B}^{0}

(If stirred continuously )

P_{T}= P_{A}^{0}\or\ P_{B}^{0}

(Layer which is above )

 

-

 

 

For two immiscible liquid;

P_{A}^{0}=P_{total}-P_{H_{2}O}^{0}=748-648\Rightarrow 100

\frac{W_{A}}{W_{B}}=\frac{P_{A}^{0}M_{A}}{P_{B}^{0}M_{B}};M_{A}=\frac{1.25}{1}\times \frac{648\times 18}{100}\Rightarrow 145.8

 


Option 1)

7.975 g/mol

Option 2)

166 g/mol

Option 3)

145.8 g/mol

Option 4)

None \: of \: these

Posted by

satyajeet

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

JCECEB 2025 registration for BTech admission through JEE Main rank begins jceceb.jharkhand.gov.in
anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

Latest Question