Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: STATEMENT - 1 : The molecular mass of acetic acid determined by depression in freezing point method in benzene and water was found to be differrent. STATEMENT - 2 : Water is polar and benzene is non-polar.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.

Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).

1.	Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: STATEMENT - 1 : The molecular mass of acetic acid determined by depression in freezing point method in benzene and water was found to be differrent. STATEMENT - 2 : Water is polar and benzene is non-polar.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - A Depression in freezing point is a colligative property which depends upon the number of particles are different in case of benzene and water that is why molecular weight of acetic acid determined by depression in freezing point method is also different.

Discussion

No Comment Found

Related InterviewSolutions

`K_(f)` for water is `1.86 K kg mol^(-1)`. IF your automobile radiator holds `1.0 kg` of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?A. 27 gB. 72 gC. 93 gD. 39 g
For a dilute solution containing `2.5 g` of a non-volatile non-electrolyte solution in `100g` of water, the elevation in boiling point at `1` atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of `Hg)` of the solution is: (take `k_(b) = 0.76 K kg mol^(-1))`A. 724B. 740C. 736D. 718
29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCIA. 2 mLB. 4 mLC. 8 mLD. 6 mL
The freezing point among the following equimolal aqueous solutions will be highest forA. `C_(6)H_(5)NH_(3)Cl` (aniline hydrochloride)B. `Ca(NO_(3))_(2)`C. `La(NO_(3))_(3)`D. `C_(6)H_(12)O_(6)` (glucose)
The amount of urea to be dissolved in 500 ml of water (K = 18.6 K `"mole"^(-1)` in 100 g solvent) to produce a depression of `0.186^(@)C` in freezing point isA. 9 gB. 6 gC. 3 gD. 0.3 g
The normality of `0.3 M` phosphorous acid `(H_(3) PO_(3))` isA. `0.1`B. `0.9`C. `0.3`D. `0.6`
Benzoic acid dissolved in benzene showsA. its normal molecular massB. Double of its normal molecular massC. Half of its normal molecular massD. Not definite.
When benzoic acid dissolve in benzene, the observed molecular mass isA. 244B. 61C. 366D. 122
Four solutions of `K_(2)So_(4)` with the concentrations 0.1m ,0.001 m, and 0.0001 mare available . The maximum value of colligative property corresponds to :A. 0.0001 m solutionB. 0.001 m solutionC. 0.01 m solutionD. 0.1 m solution
Dry air was passed successively through a solution of 5 gm of a solute in 80 gm of water and then through pure ware. The loss in weight of solution was 2.50 gm and that of pure solvent 0.04 gm. What is the molecular weight of the soluteA. `70.31`B. `7.143`C. `714.3`D. 80

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment