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  1. Problems
  2. /A solution of glucose has 22.5 g in each litre…
MediumChemistryHigh School (11–12)Solutions › Colligative Properties and the van 't Hoff FactorNumericalNEETAPAEEE (Amrita)AMUEEEAP EAPCETBCECE
+16 more exams
COMEDK UGETCUET UGGUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEESRMJEEETS EAPCETUPESEAT

A solution of glucose has 22.5 g in each litre…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solution of glucose has 22.5 g22.5\ \text{g}22.5 g in each litre. A solution of a non-electrolyte X has 3.4 g3.4\ \text{g}3.4 g in 400 mL400\ \text{mL}400 mL. At the same temperature the two solutions are isotonic. Find the molar mass of X. The molar mass of glucose is 180 g/mol180\ \text{g/mol}180 g/mol.

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Work it out on paper, then type your number. The answer itself stays hidden.

mol/L
g/mol
Need a reminder? The key ideaShowHide

The osmotic pressure of a dilute solution is π=CRT\pi = CRTπ=CRT, with CCC in mol/L and TTT in kelvin. R=0.0821 L atm K−1mol−1R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}R=0.0821 L atm K−1mol−1 gives atm, and R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}R=8.314 J K−1mol−1 gives kPa. Isotonic solutions have equal π\piπ. A column of height hhh of a liquid of density ρ\rhoρ gives a pressure ρgh\rho g hρgh.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Isotonic solutions have equal π\piπ. At the same TTT and with π=CRT\pi = CRTπ=CRT, the concentrations CCC are equal.

Step 1. Glucose. C=22.5 g/L180 g/mol=0.125 mol/LC = \dfrac{22.5\ \text{g/L}}{180\ \text{g/mol}} = 0.125\ \text{mol/L}C=180 g/mol22.5 g/L​=0.125 mol/L.

Step 2. X. The moles of X in 0.400 L0.400\ \text{L}0.400 L must give the same concentration: n=0.125×0.400=0.0500 moln = 0.125 \times 0.400 = 0.0500\ \text{mol}n=0.125×0.400=0.0500 mol.

Step 3. Molar mass. M=3.4 g0.0500 mol=68 g/molM = \dfrac{3.4\ \text{g}}{0.0500\ \text{mol}} = 68\ \text{g/mol}M=0.0500 mol3.4 g​=68 g/mol.

Answer: 0.125 mol/L0.125\ \text{mol/L}0.125 mol/L for glucose, and 68 g/mol68\ \text{g/mol}68 g/mol for X.

Check. The temperature and the gas constant cancel, so we did not need them.

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  1. Osmosis and osmotic pressureColligative Properties

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