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Solutions — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions5 Marks
Question
  1. Define the following terms:
  1. Molarity.
  2. Molal elevation constant(Kb).
  1. A solution containing 15 g urea (molar mass = $60 g\ mol^{–1}$​​​​​​​) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = $180 g\ mol^{–1}​​​​​​​$) in water. Calculate the mass of glucose present in one litre of its solution.
✓

Answer

  1.  
  1. Molarity is defined as number of moles of solute dissolved in one litre of solution.
  2. It is equal to elevation in boiling point of 1 molal solution.
  1. For isotonic solutions: $\pi$ urea = $\pi$ glucose
$\frac{\text{W}_\text{urea}}{\text{M}_\text{urea}\times\text{V}_{s}}=\frac{\text{W}_{Glucose}}{\text{M}_{Glucose}\times\text{V}_{s}}$ (As volume of solution is same)

$ \frac{\text{W}_{urea}}{\text{M}_{urea}}=\frac{\text{W}_{Glucose}}{\text{M}_{Glucose}}$ or $ \frac{\text{15g}}{\text{60g mol}^{-1}}=\frac{\text{W}_{Glucose}}{\text{180g mol}^{-1}}$

$\text{W}_{Glucose}=\frac{\text{15g}\times\text{180g mol}^{-1}}{\text{60g mol}^{-1}}=\text{45g}$

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