How many grams of Carbon are there in 1.95 moles of Glucose (C6H12O6)?
1 answer:
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Answer:
760 uM
Explanation:
<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. 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 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>
The problem here is that the amount of magnesium ion remains the same irrespective of the volume.
Amount of magnesium in the growth medium = <em>molarity x volume</em>
= 97 x x 47 x
= 4.559 x
Then, the volume reduced to 6.0 mL, the new molarity becomes;
<em>molarity = mole/volume </em>
= 4.559 x /6 x
= 7.598333 x
M = 759.83333 uM
To the correct number of significant digits = 760 uM