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Sphinxa [80]
3 years ago
14

When the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solu

tion level rises. Based on this information, what can be said about these two solutions
Chemistry
1 answer:
gtnhenbr [62]3 years ago
3 0

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.

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Find the mass number by adding 22 and 34 = 56.

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When two gases or two liquids form a solution, the substance that is present in the largest amount is the?
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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
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<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

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