Calculate the molality of a 1.52 M sugar (C_12H_22O_11) solution with a density

Question

Calculate the molality of a 1.52 M sugar (C_12H_22O_11) solution with a density of 1.12 g/mL a) 1.74 m b) 1.52 m c) 1.36 m d) 2.53 m e) none of the above A 0.86% m/m solution of NaCl is called "physiological saline" because its osmotic pressure is equal to that of the solution in blood cells. Calculate the osmotic pressure of solution at normal body temperature (37 degree C). Note that the density of the saline solution is 1.005 g/mL, and use the theoretical value of the Van't Hoff factor for NaCl a) 7.5 atm b) 3.7 atm c) 1.5 atm d) 4.7 e) none of the above

Explanation / Answer

(11)

1.52 M means 1.52 mol of solute in 1 L (1000 mL ) of solution

So, mass of solute = moles * molar mass = 1.52 * 342 = 520. g.

Given that density of solution = 1.12 g/mL

It means, mass of 1 mL of solution = 1.12 g.

Then, mass of 1000 mL of solution = 1.12 * 1000 = 1120 g.

Hence, mass of solvent = mass of solution - mass of solute = 1120 - 520. = 600 g. = 0.600 kg.

Therefore, molality = moles of solute / mass of solvent in kg.

m = 1.52 / 0.6

m = 2.53 m

(d)

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