Answer
(a) 1.7 m
(b) 0.87 m
(c) 7.0 m
Work Step by Step
(a)
1. Find the mass of solvent, considering 1 liter of solution:
- Total mass of solution:
$$1 L \times \frac{1.12 \ g}{1 \ mL} \times \frac{1000 \ mL}{1 \ L} = 1120 \ g \ sol.$$
- Mass of sugar:
$ C_{12}H_{22}O_{11} $ : ( 1.008 $\times$ 22 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 11 )= 342.30 g/mol
$$1 \ L \times \frac{1.22 \ moles}{1 \ L} \times \frac{342.30 \ g}{1 \ mol} = 418 \ g \ C_{12}H_{22}O_{11}$$
Mass of solvent = 1120 g - 418 g = 702 g,
702 g = 0.702 kg
2. Using that information, calculate the molality.
$$\frac{1.22 \ moles}{0.702 \ kg} = 1.7 \ m$$
(b)
1. Find the mass of solvent, considering 1 liter of solution:
- Total mass of solution:
$$1 L \times \frac{1.04 \ g}{1 \ mL} \times \frac{1000 \ mL}{1 \ L} = 1040 \ g \ sol.$$
- Mass of NaOH:
$ NaOH $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol
$$1 \ L \times \frac{0.87 \ moles}{1 \ L} \times \frac{40.00 \ g}{1 \ mol} = 34.8 \ g \ NaOH$$
Mass of solvent = 1040 g - 34.8 g = 1005.2 g,
1005.2 g = 1.0052 kg
2. Using that information, calculate the molality.
$$\frac{0.87 \ moles}{1.0052 \ kg} = 0.87 \ m$$
(c)
1. Find the mass of solvent, considering 1 liter of solution:
- Total mass of solution:
$$1 L \times \frac{1.19 \ g}{1 \ mL} \times \frac{1000 \ mL}{1 \ L} = 1190 \ g \ sol.$$
- Mass of NaHCO_3:
$ NaHCO_3 $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 84.01 g/mol
$$1 \ L \times \frac{5.24 \ moles}{1 \ L} \times \frac{84.01 \ g}{1 \ mol} = 440.21 \ g \ NaHCO_3$$
Mass of solvent = 1190 g - 440.21 g = 749.79 g,
749.79 g = 0.74979 kg
2. Using that information, calculate the molality.
$$\frac{5.24 \ moles}{0.74979 \ kg} = 7.0 \ m$$
