Question

A biochemist carefully measures the molarity of glycerol in 913 mL of photobacterium cell growth medium to be 81.3 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 11.1 mL.
a. Calculate the new molarity of glycerol in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.

Answer 1

Answer : The new molarity of glycerol in the photobacterium cell growth medium is, $6.69\times 10^3\mu M$

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of glycerol.

$M_2\text{ and }V_2$ are the new molarity and volume of glycerol .

We are given:

$M_1=81.3\mu M\\V_1=913mL\\M_2=?\\V_2=11.1mL$

Putting values in above equation, we get:

$81.3\mu M\times 913mL=M_2\times 11.1mL\\\\M_2=6.69\times 10^3\mu M$

Hence, the new molarity of glycerol in the photobacterium cell growth medium is, $6.69\times 10^3\mu M$