Answer:
[tex]\boxed{\text{A1: 12.5 g/L ; A2: 133 g/L; B1: 20 \% by volume; B2: 6 mL}}[/tex]
Explanation:
A. Concentration
Question 1
[tex]\begin{array}{rcl}\text{Concentration} & = & \dfrac{\text{Mass of solute}}{\text{Volume of solution}}\\\\& = & \dfrac{\text{25 g}}{\text{2 L}}\\\\& = &\textbf{12.5 g/L}\\\end{array}\\\text{The concentration of sugar is $\boxed{\textbf{12.5 g/L}}$}[/tex]
Question 2
[tex]\begin{array}{rcl}\text{Concentration} & = & \dfrac{\text{100 g}}{\text{0.750 L}}\\\\& = &\textbf{133 g/L}\\\end{array}\\\text{The concentration of the additive is $\boxed{\textbf{133 g/L}}$}[/tex]
B. Percent by volume
Question 1
The formula for percent by volume is
[tex]\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%[/tex]
If your solution contains 50 mL of rubbing alcohol and 200 mL of water, the volume of solution is 250 mL,
[tex]\begin{array}{rcl}\\\text{Percent by volume} & = & \dfrac{\text{50 mL}}{\text{250 mL}}\times 100 \, \%\\\\& = & 20 \, \%\\\end{array}\\\text{The concentration of the solution is $\large \boxed{\textbf{20 \% by volume}}$}[/tex]
Question 2
If you have 300 mL of a solution that is 2 % v/v H₂O₂,
[tex]\begin{array}{rcl}2 \, \% & = & \dfrac{\text{Volume of H$_{2}$O}_{2}}{\text{300 mL}}\times 100 \, \%\\\\2 \times \text{300 mL} & = & \text{Volume of H$_{2}$O}_{2} \times 100\\\\\text{Volume of H$_{2}$O}_{2} & = & \dfrac{2 \times 300\text{ mL}}{100}\\\\ & = & \textbf{6 mL}\\\end{array}\\\text{The volume of H$_{2}$O$_{2}$ in the solution is $\large \boxed{\textbf{6 mL}}$}[/tex]
