1. Define the following terms: 1. Molarity. 2. Molal elevation constant(Kb). 2. 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.

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. 2. For isotonic solutions: urea = glucose W_urea M_urea _ s = W_ Glucose M_ Glucose _ s (As volume of solution is same) W_ urea M_ urea = W_ Glucose M_ Glucose or 15g 60g mol^ -1 = W_ Glucose 180g mol^ -1 W_ Glucose = 15g 180g mol^ -1 60g mol^ -1 =45g

1. Define the following terms: 1. Molarity. 2. Molal elevation co…

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Question

Define the following terms:

Molarity.
Molal elevation constant$(Kb).$

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.

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Answer

Molarity is defined as number of moles of solute dissolved in one litre of solution.
It is equal to elevation in boiling point of $1$ molal solution.

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