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3 years ago

If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?

Chemistry

2 answers:

expeople1 [14]3 years ago

4 0

Answer: The cell will shrink when kept in 1% NaCl solution.

Explanation: This is explained with the help of osmosis.

Osmosis is a process in which the solvent flow from a solution of low concentration to a solution of high concentration through a semi-permeable membrane.

We a given a cell which is isotonic to 0.88% NaCl solution.

When the same cell is kept in a solution of 1% NaCl solution, osmosis will occur from inside to outside because the concentration is low inside the cell than the outside. So, the solvent will flow from inside to outside, which results in the shrinkage of cell. The solutions which have high concentration outside the cell are known as hypertonic solutions.

EleoNora [17]3 years ago

3 0

The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to extracellular fluid and equilibrium is reached.

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hichkok12 [17]

Answer:

$\Delta G^0_{rxn}=16.81kJ$

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In this case, since the Gibbs free energy of any process is related with the enthalpy change, temperature and entropy change as shown below:

$\Delta G^0=\Delta H^0-T\Delta S^0$

For a chemical reaction it is simply modified to:

$\Delta G^0_{rxn}=\Delta H^0_{rxn}-T\Delta S^0_{rxn}$

Thus, since the enthalpy of reaction is given as -304.2 kJ and the entropy as -414.2 J/K (-0.4142 kJ/K), at 775 K the Gibbs free energy of reaction turns out:

$\Delta G^0_{rxn}=-304kJ-775K*(-0.4142\frac{kJ}{K} )\\\\\Delta G^0_{rxn}=16.81kJ$

Whose result means this is a nonspontaneous reaction.

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