Answer
(a) 0.0610 m sucrose.
(b) 2.04 m ethylene glicol.
Work Step by Step
(a)
685 g = 0.685 kg
$ C_{12}H_{22}O_{11} $ : ( 1.008 $\times$ 22 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 11 )= 342.30 g/mol
- Calculate the amount of moles:
$$ 14.3 \space g \times \frac{1 \space mol}{ 342.30 \space g} = 0.0418 \space mol$$
- Calculate the molality:
$$ \frac{ 0.0418 \space mol}{ 0.685 \space kg} = 0.0610 \space m $$
(b)
3505 g = 3.505 kg
- Calculate the molality:
$$ \frac{7.15 \space mol}{ 3.505 \space kg} = 2.04 \space m $$
