Question

E connect CHM M: CHEMISTRY onnect Ch 13 Question 6 (of 11 3. value 10.00 points 3 attempts ten Check my work Be sure to answer all parts. Calculate the molality of each of the following aqueous solutions. (a) 2.55 M NaCl solution density of solution - 1.08 g/mL) (2.74 m (b) 42.9 percent by mass KBr solution 6.57 References Multipart Answer Difficulty: 2 0

Answer 1

we know, $molality = \frac{moles solute}{kg solvent}$

Let's assume 1000 mL of solution. $\therefore$ mass of solution = (1000 mL * 1.08 g/mL) = 1080 g

Moles of NaCl = 2.55

a. mass of water = 1080 g - (2.55 moles * 58.44 g/mol)NaCl = 930.978 gm = 0.931 kg

$\therefore$  

2.55moleNaCl molality = 0.931kg H2O 2.74mol/kg

b. 100 g of the solution contains 42.9 g KBr and (100 - 12.9)g = 57.1 g H2O.

mols of KBr = $\frac{42.9 g KBr}{119 g/mol KBr} =$ 0.360 mol KBr

mass of H2O (in kg) = (54.8 gm / 1000 gm) = 0.0548 kg

   $\therefore$   $molality = \frac{0.36 mol KBr}{0.0548 kg H_{2}O } = 6.57 mol/kg$