SOLUTIONS QUIZ-9

Q1. The normality of a 100 mL solution of sodium hydroxide which contains 4 g of NaOH, is

•  2
•  1.5
•  .5
•  1.0

Solution
N=(w×1000)/(eq.wt.×V(mL))=(4×1000)/(40×100)=1.0 N

Q2.The amount of ice that will separate out on cooling a solute containing 50 g of ethylene glycol in 200 g water to -9.3℃ will be

•  3.87 g
•  38.7 g
•  161.3 g
•  8.37 g

Solution
=(1000×k_f×w)/(∆T×W) 62=(1000×1.86×50)/(9.3×W) W=161.3 g Total water =200 g Hence, ice separated=(200-161.3)g=38.7 g

Q3.  Among the following substances, the lowest vapour pressure is exerted by :

•   Water
•  Rectified spirit
•  Kerosene
•  Mercury

Solution
Hg has higher attractive forces among molecules

Q4. A solution has an osmotic pressure of 0.821 atom at 300 K. its concentration would be :

•  0.066 M
•  0.66 M
•  0.33 M
•  0.033 M

Solution
πV=nST or π=cST ∴ c=0.821/(0.0821 ×300)=0.033 M

Q5.The vant’s Hoff factor for 0.1 M Ba(NO_3 )_2 solution is 2.74. The degree of dissociation is

•  87%
•  91.3%
•  100%
•  74%

Solution
 

.

Q6. What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised? (k_f for water=1.86 K mol^(-1))

•  0.85℃
•  -3.53℃
• 0℃
•  -0.35℃

Solution
∆T_f=i×k_f×m i for HBr=1 +α where, α =degree of dissociation i=1+0.9=1.9 ∴ ∆T_f= 1.9 ×1.86×(8.1×1000)/(100×81) =3.534℃ Freezing point =-3.534℃

Q7.The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is

•  0.1 M NaCl
•  0.1 M KCl
•  0.1 M BaCl_2
•  0.1 M sucrose

Solution
Colligative properties depend upon number of particles in solution and concentration of solution. Larger the number of particles in solution, higher is the colligative properties. Hence, highest boiling point is found in 0.1 M BaCl_2. BaCl_2 (aq)⟶B^(2+)+2Cl^-

Q8. 5 L of a solution contains 25 mg of Ca〖CO〗_3. What is its concentration in ppm? (mol. wt. of Ca〖CO〗_3 is 100)

•  1
•  5
•  2500
•  25

Solution
ppm =(weight of solute ×〖10〗^6)/(weight of solution ) = (25×〖10〗^(-3)×〖10〗^6)/5000 = 5

Q9.A 0.025 M solution of monobasic acid had a freezing point of -0.060℃. The pK_a for the acid is

•  1.2
•  2
•  5.7
•  2.5

Solution

Q10. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216℃ than that of the pure solvent. The molecular weight of the substance (molal elevation constant for the solvent is 2.16℃) is

•  10.1
•  100
•  10
• 1.001

Solution