Question

100 Mixtures and Solutions 90 Instructions: Use the following image to answer the NaNO following questions. 70 1. A solution contains 40 g of NaCl per 100g of water 60 CaCl at 100 C. Is the solution unsaturated, saturated or 50 POINO KC supersaturated? 9 40 NaC 30 20 KOO 2. A solution contains 50 g of KCl per 100 g of water at 25 C. Is the solution unsaturated, saturated or 10 Ce(SO); 0 10 20 30 40 50 60 70 80 90 100 supersaturated? Temperature (C) Solubility Graph (Online Image]. Retrieved November 12, 2018 from https//socratic.org/chemistry/solutions and-their-behavior/solubility-graphs A solution contains 10 g of KNO3 per 100 g of water at 30 C. Is the solution unsaturated 3. saturated or supersaturated? 4. An 8 oz bottle of Dr. Pepper made with imperial cane sugar (C12H22011) contains 27 g of in 236.6 mL of water. What is the molarity of the solution with respect to cane sugar? sugar 5. Calculate the mass of NaCl in a 55 mL sample of a 3.5 M saline solution. 6. What volume of water should be added to 2.6 g of KCI to make a 4 M solution? Solubility (g of salt in 100 g H,O) ONY

Answer 1

1) From the graph, we can see that the solubility of NaCl at 100°C is 40 grams in 100 grams of water. This is exactly the amount given in the question, so the solution is saturated (this means it has the maximum soluble concentration).

2) From the graph, we can see that the solubility of HCl at 25°C is approximately 35 g/100g. If we have 50 g/100g, this means we are over the solubility limit. This is a supersaturated solution.

3) For KNO₃, tho solubility at 30°C is approximately 45 g/100g. If there are 10 g/100g, we are below the solubility limit, so the solution is unsaturated.

4) The molarity can be calculated as:

$M=\frac{\frac{m}{m_{molar}}}{V(L)}$

The m_molar in this case is 342.3 g/mol, the m is 27 g and the volume (in units of liters) is 0.2366 L: we add this to our formula:

$M=\frac{\frac{27g}{342.3g/mol}}{0.2366L}=0.33M$

5) The mass can be calculated by the same formula we used in item 4. We re-arrange it to calculate the mass (the molar mass is 58.4 g/mol):

$m=M\cdot V(L)\cdot m_{molar}=3.5\frac{mol}{L}\cdot 58.4\frac{g}{mol}\cdot 0.055L=11.24g$

6) Yet another rearrangement of the same formula:

$V(L)=\frac{m}{M\cdot m_{molar}}=\frac{2.6g}{74.6\frac{g}{mol}\cdot 4\frac{mol}{L}}=0.00871L=8.71mL$