Calculate the mass of solute in each of the following: a. 85.0 mL of 4.5% (m/v) sucrose solution b. 275 mL of 10.5% (m/v) NaOH solution c. 1.80 L of 22.0% (m/v) LiNO3 solution.
The Correct Answer and Explanation is:
a. 85.0 mL of 4.5% (m/v) sucrose solution
Mass of sucrose=(4.5100)×85.0=3.825 g\text{Mass of sucrose} = \left( \frac{4.5}{100} \right) \times 85.0 = 3.825\ \text{g}Mass of sucrose=(1004.5)×85.0=3.825 g
b. 275 mL of 10.5% (m/v) NaOH solution
Mass of NaOH=(10.5100)×275=28.875 g\text{Mass of NaOH} = \left( \frac{10.5}{100} \right) \times 275 = 28.875\ \text{g}Mass of NaOH=(10010.5)×275=28.875 g
c. 1.80 L of 22.0% (m/v) LiNO₃ solution
First convert 1.80 L to mL: 1.80 L=1800 mL1.80\ \text{L} = 1800\ \text{mL}1.80 L=1800 mL
Now calculate: Mass of LiNO₃=(22.0100)×1800=396.0 g\text{Mass of LiNO₃} = \left( \frac{22.0}{100} \right) \times 1800 = 396.0\ \text{g}Mass of LiNO₃=(10022.0)×1800=396.0 g
Explanation (300 words style):
In chemistry, a mass/volume percent solution (% m/v) expresses the concentration of a solute in a given volume of solution. Specifically, it is defined as the number of grams of solute dissolved in 100 milliliters (mL) of solution. This is commonly used in biological and medical sciences due to its practical convenience when measuring liquids and solids.
To find the mass of solute in a solution with a known % (m/v) and volume, we rearrange the definition: \text{% (m/v)} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \right) \times 100
Solving for mass: \text{mass of solute} = \left( \frac{\text{% (m/v)}}{100} \right) \times \text{volume (mL)}
In part (a), we had 85.0 mL of a 4.5% sucrose solution. This means every 100 mL contains 4.5 g of sucrose. Therefore, 85 mL contains proportionally less, yielding 3.825 g of sucrose.
In part (b), the solution has a higher concentration (10.5%), and a larger volume (275 mL), so the amount of solute, NaOH, is also higher at 28.875 g.
In part (c), we convert 1.80 L to 1800 mL to stay consistent with units. At 22.0%, the solution is quite concentrated, and the larger volume results in a significant 396.0 g of lithium nitrate.
Understanding this method allows chemists and health professionals to prepare solutions with precise concentrations for lab experiments, medical treatments, or industrial processes.
