4) elevation in boiling point is a colligative property which depends on solute concentration. More number of solute particles means more increase in boiling point ( or lesser freezing point ) . Effective concentration is calculated by multiplying van’ t Hoff factor ( i ) with concentration c.
Van’t Hoff factor is number of particles after dissociation or association for eg vant Hoff factor (i ) for NaCl = 2
( a ) 1 M NaCl
i = 2 & c = 1
i x c = 2
( b ) 1M C6H12O6 ( it is non electrolyte )
i = 1 & c = 1
i x c = 1
( c ) 1 M NaNO3
i = 2 & c = 1
i x c = 2
( d ) 1 M CaCl2
i = 3 & c = 1
i x c = 3
( e ) 1 M Ca( NO3 )2
i = 3 & c= 1
i x c = 3
( f ) 2 M C6H12O6
i x c = 2
( g ) 2 M NaCl
i x c = 2 x 2 = 4
( h ) 2 M CaCl2
i x c = 3 x 2 = 6
boiling point order
1M C6H12O6 < 2 M C6H12O6 = 1 M NaCl = 1 M NaNO3 < 1 M CaCl2 < 1M Ca( NO3 )2 < 2M NaCl < 2MCaCl2 ( reverse order for freezing point )
4. Rank the following solutions by increasing boiling point (or decreasing freezing point). Note: Some solutions ma...
4. Rank the following solutions by increasing boiling point (or decreasing freezing point). Note: Some solutions may have the same boiling points. Indicate when boiling points are the same 1.0 M NaCl, 1.0 M CH20., 1.0 M NaNO3, 1.0M CaCl, 10 M Ca(NO3)2, 2.0 M C6H12O6, 2.0 M NaCl, 2.0 M CaCl2
24. Circle one word inside each set of parentheses in the following statement: (Solutions, colloids, suspensions) contain the smallest particle size and (solutions, colloids, suspensions) contain the largest particle size. Which statement about colligative properties is incorrect? A. Solutes will lower the freezing point and raise the boiling point of a solvent. B. Colligative properties depend only on the number of dissolved particles and not on their size or chemical properties. C. Colligative properties involve only charged particles in solution....
Doblydorlog on to com 2. Rank order the following aqueous solutions in order of increasing boiling point elevation. 0.100 mol/kg HBr 3.1.2 5.2 ЈТве - Комі 0.100 mol/kg MgSO4 =. 0.010 mol/kg NaCl lowing 8.00 a 0.010 mol/kg glucose 0.010 mol/kg CaCl2 A. HBr <MgSO4 = CaCl2<NaCl < glucose B. MgSO4 < HBr < CaCl2 < NaCl <glucose C.) glucose < NaCl <CaCl2 <HBr <MgSO4 D. glucose < CaCl2< NaCl <HBr <MgSO4 E. glucose < CaCl2< NaCl <MgSO4 = HBr...
Arrange the following aqueous solutions in order of increasing boiling point: 0.25 m NaCl (sodium chloride) 0.15 m CaBr2 (calcium bromide) 0.4 m C6H12O6 (sucrose) 0.25 m HCH3COO (acetic acid) Explain the reasons you ordered them as you did.
Which of the following aqueous solutions has the lowest freezing point? The Kf of water is 1.86 kg.oC/mol. a. 0.8 m CH3OH (methanol) b. 0.4 m NaCl c. 0.3 m MgBr2 d. 0.2 m AlF3 e. 0.25 m Ca(NO3)2 Calculate the freezing point ONLY for the solution you marked as your answer above.
List the following aqueous solutions in order of decreasing freezing point: 0.13 M NaCl, 0.36 M NH3 , 0.12 M NiCl2, 0.12 M NH3.
Order the following solutions in ascending order of their boiling point: 0.0825 m NaCl, 0.050 m KCl, 0.050 M Na2SO4, 0.055m Ca (NO3) 2, 0.050 m Na3PO4. Determine the osmotic pressure of the solutions.
Which of the following aqueous solutions has the highest freezing point? C. is correct, can you explain why? A. 0.12 m (NH4)3PO4 B. 0.25 m Ca(NO3)2 C. Correct 0.15 m MgCl2 D. 0.25 m NaCl
Arrange the following aqueous solutions in order of increasing boiling point temperature (lowest to highest temperature, with 1 being the lowest and 4 being the highest): 0.20 m glucose, 0.30 m BaCl2, 0.40 m NaCl, 0.50 m Na2SO4. 0.20 m glucose [ Choose ] 2 3 1 4 0.30 m BaCl2 [ Choose ] 2 3 ...
Question 11 (4 points) Which of the following aqueous solutions would have the highest boiling point? 0 1.0 mole BaCl2 in 1.0 kg of water 01.0 mole NaCl in 1.0 kg of water O 1.0 mole NH4Cl in 1.0 kg of water O 1.0 mole Na3PO4 in 1.0 kg of water 1.0 mole LINO3 in 1.0 kg of water