Identify the correct statementA. 1 M NaCl has higher freezing point than 1 M glucose solutionB. 1 M NaCl solution has same boiling point as 1 M KCl solutionC. Molecular weight NaCl will be less than 58.5 in water because it undergoes diffociationD. `i gt 1` when solute undergoes association

Identify the correct statementA. 1 M NaCl has higher freezing point th

1.	Identify the correct statementA. 1 M NaCl has higher freezing point than 1 M glucose solutionB. 1 M NaCl solution has same boiling point as 1 M KCl solutionC. Molecular weight NaCl will be less than 58.5 in water because it undergoes diffociationD. `i gt 1` when solute undergoes association
Answer» Correct Answer - B::C Explanation : (b) is correct both have i=2 and isotonic (same) molar conc. (c ) Is correct because `NaCl rarr Na^(+)+Cl^(-)`, molecular weight will be less than 58.5. (a) Is wrong because non-volatile solute lowers down the freezing point. (d) Is not correct because ilt1, when association takes place.

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

Identify the correct statementA. 1 M NaCl has higher freezing point than 1 M glucose solutionB. 1 M NaCl solution has same boiling point as 1 M KCl solutionC. Molecular weight NaCl will be less than 58.5 in water because it undergoes diffociationD. `i gt 1` when solute undergoes association

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment