Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
For Business Enquiry

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

Ask Solutions question free ×

Queries asked on Sunday and after 7 pm from Monday to Saturday will be answered after 12 pm the next working day.

Solutions Free Doubts and Solutions

NEET - NEET - Chemistry - Solutions

A solution M is prepared by mixing ethanol and H2O.The mole fraction of ethanol in mixture is 0.9     . Given: Kf H20-1.86 Kkg/mol ,Kf ethanol-2.0 Kkg/mol ,Kb H2O-0.52 Kkg/mol ,Kb ethanol-1.2 Kkg/mol, Standard freezing point of H20- 273K , Standard freezing point of ethanol- 155.7K , Standard boiling point of H2O- 373K , Standard boiling point of ethanol-351.5K ,vp of pure water -32.8mm Hg,vp of pure ethanol -40mm Hg, mol wt of H2O - 18g/mol, mol wt of ethanol- 46g/mol                                . Consider the solutions to be ideal dilute solutions and solutes to be non volatile and non-dissociative                                                                                                                                        .Ques 1 ) The fp of solution M is: (a) 268.7K ,(b) 268.5 K ,(c) 234.2 K , (d) 150.9 K                          .Ques 2 ) The vp of solution M is: (a) 39.3mmHg ,(b) 36.0mmHg ,(c) 39.5mmHg ,(d) 28.8mmHg    .Ques 3) Wter is added to the solution M such that the mole fraction of water in solution becomes 0.9. The bp of this solution is : (a)380.4K ,(b)376.2K, (c)375.5K ,(d)354.7K

Asked by Balbir 30th July 2019, 6:16 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

What is the composition of the vapour which is in equilibrium at 30 degree celcius with a benzene-toluene solution with a mole fraction of benzene of 0.400? (P0 B is 119 torr and P0 T is 37 torr)       (a) 1.237  (b) 2.237  (c) 3.237  (d) 0.237

Asked by Balbir 28th July 2019, 7:59 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

By dissolving 5g substance in 50g of water, the decrease in freezing point is 1.2 degree celcius. The gram molal depression is 1.85 degree celcius. The molecular weight of substance is (a)105.4 (b)118.2 (c)137.2 (d)154.2  

Asked by Balbir 28th July 2019, 6:08 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

Calculate the normality of 250 mL aqueous solution of H2SO4 having pH -0.0 (a)0.25 N (b)0.50 N (c)1 N (d)2 N

Asked by Balbir 9th July 2019, 10:25 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

Pls solving the follin

qsnImg
Asked by ntg432000 28th April 2019, 4:25 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

What is the [OH-] in the final solution prepared by mixing 20 mL of 0.050 M HCl with 30 mL of 0.10 M Ba(OH)2 ?

Asked by kumarisakshi0209 22nd March 2019, 11:41 AM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

The osmotic pressure of 5% aqueous soln of sugar (mol mass 342) at 15 degree Celsius is? 

Asked by Sri.prabhavani 15th March 2019, 12:31 AM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET - NEET - Chemistry - Solutions

Please answer the following question with explanation

qsnImg
Asked by deepakudgiri29 7th January 2019, 9:56 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

NEET1 - NEET1 - Chemistry - Solutions

18%(w/V) solution of urea (mol.wt=60) is

Asked by priya47 28th September 2018, 6:12 PM
Answered by Expert

Answer this question

×
Your answer has been posted successfully!

Latest Questions

Chat with us on WhatsApp