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If the osmotic pressure of a 0.010 M aqueous solution of sucrose at 27˚C is 0.25 atmosphere, then the osmotic pressure of a 0.010 M aqueous solution of NaCl at 27˚C is

(a) 0.062 atm

(b) 0.12 atm

(c) 0.25 atm

(d) 0.50 atm

(e) 1.0 atm

The correct answer is (d). However, I'm not entirely sure why.

I know that the osmotic press $\Pi$ = MRT where M is the concentration, R is the ideal gas constant, and T is the absolute temperature.

Thus:

$\Pi_{NaCl} = M_{NaCl}RT = (0.01 M)RT = M_{sucrose}RT = 0.25$

So why isn't the osmotic pressure of NaCl at 27˚C 0.25 atm?

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The expression you need is

$$\Pi = i\mathrm{MRT}$$

where $i$, the van't Hoff factor, describes the degree of ionization of the solute (for sucrose, it's 1; for $\ce{NaCl}$, it's 2, since we get 2 moles of ions for each mole of salt we dissolve). The other variables are as you describe. Thus, the osmotic pressure for your salt solution is twice that of your sucrose solution.

See this page for a nice explanation and worked examples for sucrose and salt.

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