Answer
(a) 2.74 m
(b) 6.93 m
Work Step by Step
(a)
1. Find the mass of solvent, considering 1 liter of solution:
- Total mass of solution:
$$1 L \times \frac{1.08 \ g}{1 \ mL} \times \frac{1000 \ mL}{1 \ L} = 1080 \ g \ sol.$$
- Mass of $NaCl$:
$ NaCl $ : ( 22.99 $\times$ 1 )+ ( 35.45 $\times$ 1 )= 58.44 g/mol
$$1 \ L \times \frac{2.55 \ moles}{1 \ L} \times \frac{58.44 \ g}{1 \ mol} = 149 \ g \ NaCl$$
Mass of solvent = 1080 g - 149 g = 931 g,
931 g = 0.931 kg
2. Using that information, calculate the molality.
$$\frac{2.55 \ moles}{0.931 \ kg} = 2.74 \ m$$
(b)
- Considering 100 g:
45.2 g of $KBr$
100 g - 45.2 g= 54.8 g of solvent = 0.0548 kg of solvent
$ KBr $ : ( 79.90 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 119.00 g/mol
$$ 45.2 \space g \times \frac{1 \space mole}{ 119.00 \space g} = 0.380 \space mole$$
$$Molality = \frac{0.380 \ mole}{0.0548 \ kg} = 6.93 \ m$$
