Solutions

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Question
CBSEENCH12011360

0.5 Molal aqueous solution of aweak acid (HX) is 20% ionised. If Kf for water is1.86 K kg mol-1, the lowering in freezing point of the solution is:

  • -1.12 K 

  • 0.56 K 

  • 1.12 K 

  • -0.56 K 

Solution

C.

1.12 K 

WiredFaculty

Sponsor Area

Question
CBSEENCH12011403

1.00 g of a non- electrolyte solute (molar mass 250 g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K Kg mol-1, the freezing point of benzene will be lowered  by:

  • 0.4 K

  • 0.3 K

  • 0.5 K

  • 0.2 K

Solution

A.

0.4 K

Molality of non- electrolyte solute

WiredFaculty

Question
CBSEENCH12011244

200 mL of an aqueous solution of a protein contains its 1.26 g . The osmotic pressure of this solution at 300 K is found to be  2.57 x 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1)

  • 51022 g mol-1

  • 122044 g mol-1

  • 31011 g mol-1

  • 61038 g mol-1

Solution

D.

61038 g mol-1

WiredFaculty

Question
CBSEENCH12011284

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely molar concentration of sodium ion, Na+ and carbonate ion, CO32- are respectively (Molar mass of Na2CO3 = 106 g mol-1)

  • 0.955 M and 1.910 M

  • 1.910 M and 0.955 M 

  • 1.90 M and 1.910 M 

  • 0.477 M and 0.477 M 

Solution

B.

1.910 M and 0.955 M 

WiredFaculty

Question
CBSEENCH12011324

A 0.0020 m aqueous solution  of an ionic compound Co(NH3)5(NO2)Cl freezes at -0.00732o C . Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (kf = - 1.86o C/m)

  • 2

  • 3

  • 4

  • 1

Solution

A.

2

Given,
molality = 0.0020 m
Δ Tf = 0o C -0.007320 C
kf = 1.86 oC/m

ΔTf = i.kf x m
i = ΔTf/ kf x m
= 0.00732/1.82 x 0.0020
= 1.92 = 2

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