Chapter 9, Problem 1QAP

(a)

To determine

The mass of sugar dissolved in $m_w=1500\text{g}$ water at $T=90{}^{o}C$ .

Answer to Problem 1QAP

The mass of sugar dissolved in $m_w=1500\text{g}$ water at $T=90{}^{o}C$ is $m_s=5.022kg$ .

Explanation of Solution

Given Amount of water is $m_w=1500\text{g}$ .

Temperature is $T=90{}^{o}C$ .

Formula used:

  $C_{sugar}=\frac{m_{sugar}}{m_{sugar}+m_{water}}\times 100\%$

Where,

  $C_{sugar}$ is the composition of sugar in weight percent.

  $m_{sugar}$ is the mass of sugar.

  $m_{water}$ is the mass of water.

  

According to diagram $C_{sugar}=77$

  $\%wtsugar$

Calculations Put value of $C_{sugar}=77$ , mass of water= $\text{1500g}$

  $\begin{array}{c}{C}_{sugar}=\frac{m_{sugar}}{m_{sugar}+m_{water}}\times 100\%\\ 77=\frac{m_{sugar}}{m_{sugar}+1500\left(\text{kg}\right)}x100\%\\ 0.77=\frac{m_{sugar}}{m_{sugar}+1500\left(\text{kg}\right)}\\ m_{sugar}=0.77\left(m_{sugar}+1500\left(\text{kg}\right)\right)\\ 0.23m_{sugar}=1155\left(\text{kg}\right)\\ m_{sugar}=5.022\text{kg}\end{array}$

Conclusion So, the mass of sugar dissolved in $m_w=1500\text{g}$ water at $T=90{}^{o}C$ .is $m_s=5.022kg$ .

(b)

To determine

Composition of saturated liquid solution at $T=20°C$ .

Answer to Problem 1QAP

Composition of the saturated liquid solution is $C=65\%\text{wt}$ .

Explanation of Solution

From the sugar-water phase diagram when the liquid saturated solution from temperature $T=90°C$ is cooled to temperature $T=20°C$ then the composition of the saturated liquid solution is $C=65\%\text{wt}$ . It is shown in the graph below.

  

Conclusion Hence, the composition of the saturated liquid solution is $C=65\%\text{wt}$ .

(c)

To determine

Mass of solid sugar comes out of the solution upon cooling to $T=20°C$ .

Answer to Problem 1QAP

Mass of solid sugar comes out of the solution upon cooling to $T=20°C$ is $m_{solidified}=2.714\text{kg}$ .

Explanation of Solution

Given Amount of water is $m_w=1500\text{g}$ .

Cooling Temperature is $T=20°C$ .

Mass of sugar dissolved at $T\text{'}=90°C$ is $m_s=5.022\text{kg}$ .

Formula used:

Solubility of sugar in water is given by:

  ${\left({\text{C}}_{\text{sugar}}\right)}_{T=20°C}\text{=}\frac{{\text{m'}}_{\text{sugar}}}{{\text{m'}}_{\text{sugar}}{\text{+m}}_{\text{water}}}\text{x100\%}$

Where,

  $C_{sugar}$ =Composition of sugar in weight percent

  $m{\text{'}}_{sugar}$ =mass of maximum solubility of sugar at $T=20°C$ .

  $m_{water}$ =mass of water

Mass of sugar solidifies upon cooling to temperature $T=20°C$ is

  $m_{solidified}={\left(m_{sugar}\right)}_{T\text{'}=90°C}-{\left(m{\text{'}}_{sugar}\right)}_{T=20°C}$

  

According to diagram $C_{sugar}=65\%\text{wt}$ sugar

Calculations Put value of $C_{sugar}=65$ , mass of water $m_w=1500\text{g}$

  $\begin{array}{c}{\left(C_{sugar}\right)}_{T=20°C}=\frac{m{\text{'}}_{sugar}}{m{\text{'}}_{sugar}+m_{water}}\times 100\%\\ 65=\frac{m{\text{'}}_{sugar}}{m{\text{'}}_{sugar}+1500\left(\text{g}\right)}\times 100\%\\ \frac{m{\text{'}}_{sugar}}{m{\text{'}}_{sugar}+1500\left(\text{g}\right)}=0.65\\ m{\text{'}}_{sugar}=0.65\left(m{\text{'}}_{sugar}+1500\left(\text{g}\right)\right)\\ 0.35m{\text{'}}_{sugar}=975\left(\text{g}\right)\\ m{\text{'}}_{sugar}=\frac{975\left(\text{g}\right)}{0.35}\\ m{\text{'}}_{sugar}=2785.72\text{g}\\ m{\text{'}}_{sugar}=2.786\text{kg}\end{array}$

Mass of sugar solidifies upon cooling to temperature $T=20°C$ is

  $\begin{array}{c}{m}_{solidified}={\left(m_{sugar}\right)}_{T\text{'}=90°C}-{\left(m{\text{'}}_{sugar}\right)}_{T=20°C}\\ m_{solidified}=5.5\left(\text{kg}\right)-2.786\left(\text{kg}\right)\\ m_{solidified}=2.714\text{kg}\end{array}$

Conclusion So, the mass of sugar that solidified upon cooling to temperature $T=20°C$ is $m_{solidified}=2.714\text{kg}$ .