Solved Examples

Question 1: 

Arginine vasopressin is a pituitary hormone. It helps to regulate the amount of water in the body by reducing the flow of urine form kidneys. An aqueous solution containing 21.6 mg of vasopressin in 100 mL of solution had an osmotic pressure of 3.70 mm Hg at 25^oC. What is molecular weight of hormone? 

Solution:

We know πV = nRT

πV = w_B/m_B RT

m_B = w_B×RT/πV     …(i) 

where w_B = mass of solute (21.6× 10⁻³ g)

m_B = molar mass of solute 

R = 0.0821 L atm K⁻¹ mol⁻¹

T = 298 K

V = 100/1000 = 0.1 L : π = 3.70/760 atm 

From (i), m_B = 21.6×10^–3×0.0821×298/(3.70/760)×0.1 = 1085 g/mol  

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Question 2: 

A solution is prepared by dissolving 1.08 g of human serum albumin, a protein from blood plasma, in 50 cm³ of aqueous solution. The solution has an osmotic pressure of 5.85 mm Hg at 298 K

(a) What is molar mass of albumin?

(b) What is height of water column placed in solution? 

Given that density of water is 1gm cm³ .

Solution: 

(a) Molar mass of albumin cab be calculated using followed relation 

 m_B = w_B×RT/πV                                 …(i)

Given w_B = 1.08; R = 0.0821 litre atm K⁻¹ mol⁻¹ 

T = 298 K, π = 5.85/760 atm; V = 50/1000 = 0.05 litre  

Substituting these values in (i)

m_B = 1.08×0.0821×298/(5.85/760)×0.05 = 68655 g/mol   

(b) π h dg 

5.85/760 × 1010325 = h × 1 × 10^–3 × 9.8  

∴  h = 7.958 × 10⁻² m = 7.958 cm

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Question 3: 

Calculate osmotic pressure of 5% solution of cane sugar (sucrose) at 15^oC. 

Solution:  

m = mol. mass of sucrose (C₁₂H₂₂O₁₁) = 342 

w = 5 g,     

V = 100 mL = 0.1 litre 

S = 0.082,  

T = (15 + 273) = 288 K 

Applying the equation PV = w/m ST,  

  P = 5./342 × 1/0.1 × 0.082 × 288    = 3.453 atm

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Question 4: 

The solution containing 10 g of an organic compound per litre showed an osmotic pressure of 1.16 atmosphere at 0oC. Calculate the molecular mass of the compound (S = 0.0821 litre atm per degree per mol.) 

Solution:

Applying the equation 

m = w/PV . ST 

Given w = 10 g, P = 1.18 atm, V = 1 litre, S = 0.0821 and T = 273 K. 

m = 10/1.18×1 × 0.0821 × 273 = 189.94  

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Question 5: 

On dissolving 10.8 glucose (m.w. = 180) in 240 g of water, its boiling point increases by 0.13^oC. Calculate the molecular elevation constant of water.

Solution:

T = 100K'×w/W×m   

 or K' = T×W×m/100×w   

Given, T =0.13^oC, W = 240 g, m = 180 and 2 = 10.8 g   

K' = 0.13×240×180/100×10.8 = 5.2 ^oC

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Question 6:

A solution of 2.5 g of a non-volatile solid in 100 g benzene boiled at 0.42^oC higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is 2.67 K kg mol⁻¹.   

Solution:

m = 1000K_b×w/W×?T

Given, K_b = 2.67, w = 2.5 g, W = 100 g, T = 0.42   

m = 1000×2.67×2.5/100×0.42 = 158.9

The molecular mass of substance is 158.9.

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Question 7:

The molal elevation constant for water is 0.56 K kg mol⁻¹. Calculate the boiling point of a solution made bt dissolving 6.0 g of urea (NH₂CONH₂) in 200 g of water.   

Solution:  

T = 1000K_b×w/m×W

Given, K_b = 0.56 K kg mol⁻¹, w = 6.0 g, W = 200 g, m=60

T = 1000×0.56×6.0/200×60 = 0.28^oC

Thus, The boiling point of solution = b.pt. of water + T = (100^oC + 0.28^oC) = 100.28^oC  

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Question 8:

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^oC. Calculate the molecular mass of the substance. [Molal depression constant for water = 1.863 K kg mol⁻¹]  

Solution: 

m = 1000K_f× w/W×T

Given, K_f = 1.863 K kg mol⁻¹.  

 w = 13.6 g, W = 20 g, T = 3.7^oC  

m = 1000×1.863×13.6/20×3.7 = 243.39

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Question 9:

On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density 0.8 g/mL), its freezing point decreases by 0.40^oC. Calculate the molecular mass of non-volatile substance (K_f = 5.12 K kg mol⁻¹).  

Solution: 

Mass of benzene, W = volume × density  = 30 × 0.8 = 24 g  

Given, K_f = 5.12 K kg mol⁻¹, w = 0.25 g, T = 0.40^oC.  

We know that  

m = 1000K_f× w/W×T

= 1000×5.12×0.25/24×0.40 = 133.33

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Question 10:

What is the volume of solution containing 1 g mole of sugar that will give rise to an osmotic pressure of 1 atmosphere at 0^oC ?

Solution:

Applying the equation PV = nST, 

 V = n/P × S × T

Given n = 1, P = 1 atm, S = 0.821 and T = 273 K

 V = 1/1 × 0.0821 × 273 = 22.4 litre

Related Resources

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• Look here for reference books for chemistry 

• You can also refer methods of expressing concentration of solution

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