4.) What is the molarity of each of the following solutions?
a.) 3.6 g of HCl in enough water to make 500 mL of solution

b.) 160 g of NaOH in enough water to make 6.0 L of solution

5.) Suppose you have a supply of 3.0 M NaOH solution in your laboratory. You take 200mL of this solution and evaporate this to dryness. How many grams of solid do you end up with?

[tex]\begin{aligned} \text{moles of solute} & = \frac{\text{mass of solute}}{\text{molar mass of solute}} \\ & = \frac{\text{3.6 g}}{\text{36.46 g/mol}} \\ & = \text{0.09874 mol} \end{aligned}[/tex]

Final Step: Calculate the molarity of the solution.

Note that 500 mL = 0.500 L.

[tex]\begin{aligned} \text{molarity} & = \frac{\text{moles of solute}}{\text{volume of solution}} \\ & = \frac{\text{0.09874 mol}}{\text{0.500 L}} \\ & = \text{0.20 mol/L} \\ & = \boxed{0.20 \: M} \end{aligned}[/tex]

[tex]\tt{\huge{\red{4b) \: Solution:}}}[/tex]

Step 1: Calculate the number of moles of solute (NaOH).

Note that the molar mass of NaOH is 40.00 g/mol.

[tex]\begin{aligned} \text{moles of solute} & = \frac{\text{mass of solute}}{\text{molar mass of solute}} \\ & = \frac{\text{160 g}}{\text{40.00 g/mol}} \\ & = \text{4.0 mol} \end{aligned}[/tex]

Final Step: Calculate the molarity of the solution.

[tex]\begin{aligned} \text{molarity} & = \frac{\text{moles of solute}}{\text{volume of solution}} \\ & = \frac{\text{4.0 mol}}{\text{6.0 L}} \\ & = \text{0.67 mol/L} \\ & = \boxed{0.67 \: M} \end{aligned}[/tex]

[tex]\tt{\huge{\red{5) \: Solution:}}}[/tex]

Step 1: Calculate the number of moles of solute (NaOH).

Note that 200 mL = 0.200 L

[tex]\begin{aligned} \text{moles of solute} & = \text{molarity} \times \text{volume of solution} \\ & = 3.0 \: M \times \text{0.200 L} \\ & = \text{3.0 mol/L} \times \text{0.200 L} \\ & = \text{0.60 mol} \end{aligned}[/tex]

Final Step: Calculate the mass of solute.

[tex]\begin{aligned} \text{mass of solute} & = \text{moles of solute} \times \text{molar mass of solute} \\ & = \text{0.60 mol} \times \text{40.00 g/mol} \\ & = \boxed{\text{24 g}} \end{aligned}[/tex]

[tex]\\[/tex]

Note: Kindly swipe the screen to the left to see the continuation of the answer on the right side.

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