Question

a biochemist carefully measures the molarity of calcium carbonate in 39.8 ml of photobacterium cell growth medium to be 60.8 μm. unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. the volume of the cell growth medium falls to 4.40 ml. calculate the new molarity of calcium carbonate in the photobacterium cell growth medium. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Use the dilution - formula $C_1V_1 = C_2V_2$.

The initial molarity $C_1=60.8\ \mu M$, the initial volume $V_1 = 39.8\ mL$, and the final volume $V_2=(39.8 - 4.40)\ mL$.

Step2: Rearrange the formula to solve for $C_2$.

$C_2=\frac{C_1V_1}{V_2}=\frac{60.8\ \mu M\times39.8\ mL}{39.8\ mL - 4.40\ mL}=\frac{60.8\ \mu M\times39.8\ mL}{35.4\ mL}$.

Step3: Calculate the value of $C_2$.

$C_2=\frac{60.8\times39.8}{35.4}\ \mu M=\frac{2429.84}{35.4}\ \mu M\approx68.7\ \mu M$.

Answer:

$68.7\ \mu M$