Based on solute `-` solvent interactions, arrange the following in order of increasing solubility in `n-`octane and explain the result. Cyclohexane, `KCl,CH_(3)OH,CH_(3)CN`.

Based on solute `-` solvent interactions, arrange the following in ord

1.	Based on solute `-` solvent interactions, arrange the following in order of increasing solubility in `n-`octane and explain the result. Cyclohexane, `KCl,CH_(3)OH,CH_(3)CN`.
Answer» (a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions. (b) KCl is an ionic compound, KCl will not dissovle in n-octane. (c ) `CH_(3)OH` is polar. `CH_(3)OH` will dissolve in n-octane. (d) `CH_(3)CN` is polar but lesser than `CH_(3)OH`. Therefore, it will dissolve in n-octane but to a greater extent as compared to `CH_(3)OH`. Hence, the oreder is `KCl lt CH_(3)OH lt CH_(3)CN lt "Cyclohexane"`.

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

Based on solute `-` solvent interactions, arrange the following in order of increasing solubility in `n-`octane and explain the result. Cyclohexane, `KCl,CH_(3)OH,CH_(3)CN`.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment