Koncentriranje raztopin

Pripraviti želimo 500 kg 55 % raztopine. Koliko kilogramov 40 % raztopine potrebujemo?

Izpis podatkov:

$m_r(1) = 500 \text{ kg}$
$w(1) = 0,55$
$m_r(2) = ?$ 

Ker imamo v masnem deležu izraženo maso, bomo zapisali: 

$m_t(1) = m_t(2)$ ; ker je $m_t=w \cdot m_r$, sledi:
$w(1) \cdot m_r(1) = w(2) \cdot m_r(2)$
$$m_r(2) = \frac{w(1) \cdot m_r(1)}{w(2)}$$
$$m_r(2) = \frac{0,55 \cdot 500 \text{ kg}}{0,40} = 688 \text{ kg}$$

Potrebujemo 688 kg raztopine z masnim deležem 0,40.

Izpis podatkov:

$m_r(1) = 500 \text{ kg}$
$w(1) = 0,55$
$w(2) = 0,40$
$m_r(2) = ?$
$m_t(3) = ?$ 

$m_t(1) = m_t(2) + m_t(3)$ ; ker je $m_t=w \cdot m_r$, sledi:
$w(1) \cdot m_r(1) = w(2) \cdot m_r(2) + m_t(3)$; $m_r(1) = m_r(2) + m_t(3)$

$w(1) \cdot m_r(1) = w(2) \cdot (m_r(1) – m_t(3)) + m_t(3)$
$w(1) \cdot m_r(1) = w(2) \cdot m_r(1) – w(2) \cdot m_t(3) + m_t(3)$
$w(1) \cdot m_r(1) – w(2) \cdot m_r(1) = m_t(3) \cdot (1 – w(2))$

$$m_t(3) =  \frac{m_r(1) \cdot (w(1) – w(2))}{1 – w(2)} = \frac{500 \text{ kg} \cdot (0,55–0,40)}{0,60}$$

$m_t(3) =125 \text{ kg}$  

$m_r(2) = m_r(1) – m_t(3) = 500 \text{ kg} – 125 \text{ kg} = 375 \text{ kg}$ 

Potrebujemo torej 375 kg raztopine z masnim deležem 0,40 in ji moramo dodati 125 kg topljenca.