Answer:
because there more acid in diet cola
Answer:
The unknown solution had the higher concentration.
Explanation:
When two solutions are separated by a semi-permeable membrane, depending on the concentration gradient between the two solutions, there is a tendency for water molecules to move across the semi-permeable in order to establish an equilibrium concentration between the two solutions. This movement of water molecules across a semi-permeable membrane in response to a concentration gradient is known as osmosis. In osmosis, water molecules moves from a region of lower solute concentration or higher water molecules concentration to a region of higher solute concentration or lower water molecules concentration until equilibrium concentration is attained.
Based on the observation that when the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solution level rises, it means that water molecules have passed from the glucose solution through the semipermeable membrane into the unknown solution. Therefore, the solution has a higher solute concentration than the glucose solution.
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
