The solutions listed below are in two arms of a tube separated by a semipermeable membrane. The solution in the left arm is given first. In which case the osmotic flow proceeds from the right to the left arm? Group of answer choices

0.010 M NH3 (aq) |

0.010 M NaCl (aq)

0.010 M NaCl (aq) |

0.010 M FeCl3 (aq)

0.010 M NaCl (aq) |

0.010 M CH3COOH (aq)

0.010 M NaCl (aq) |

0.010 M CaCl2 (aq)

0.010 M CaCl2 (aq) |

0.010 M FeCl3 (aq)

Question

Biology

Zoey Avery

6 March, 16:32

The solutions listed below are in two arms of a tube separated by a semipermeable membrane. The solution in the left arm is given first. In which case the osmotic flow proceeds from the right to the left arm? Group of answer choices

0.010 M NH3 (aq) |

0.010 M NaCl (aq)

0.010 M NaCl (aq) |

0.010 M FeCl3 (aq)

0.010 M NaCl (aq) |

0.010 M CH3COOH (aq)

0.010 M NaCl (aq) |

0.010 M CaCl2 (aq)

0.010 M CaCl2 (aq) |

0.010 M FeCl3 (aq)

+3

Answers (1)

Know the Answer?

Edith · Answer 1

The correct answer is the couple: 0.010 M NaCl | 0.010 M CH₃COOH

Explanation:

A semipermeable membrane allows to pass through solvent molecules but not solute molecules. In this case, all solutions have the same molarity (M) but they do not have the same quantity of solute particles because some of them dissociate in two or more ions (van't Hoff factor i is higher than 1). Osmotic flow proceeds always from the side with lower concentration of solute (with more solvent molecules) to the side with higher concentration of solute. For each pair of solution, we have to determine the number of particles of solute or the van't Hoff factor (i). If the right side has the lower concentration of solute (higher i), the osmotic flow will proceed from the right to the left.

0.010 M NH3 (aq) | 0.010 M NaCl (aq) : NH₃ is a nonelectrolyte (i=1) and NaCl has i = 2. Osmotic flow proceeds from left to the right.

0.010 M NaCl (aq) | 0.010 M FeCl3 (aq) : NaCl has i=2 and FeCl₃ has i=3 (it dissociates in Fe⁺ and 3 Cl⁻). Osmotic flow proceeds from left to the right.

0.010 M NaCl (aq) | 0.010 M CH3COOH (aq) : NaCl has i=2 and CH₃COOH is a nonelectrolyte (i=1). The lower concentration is in the right side, so the osmotic flow proceeds from right to left.

0.010 M NaCl (aq) | 0.010 M CaCl2 (aq) : NaCl has i=2 and CaCl₂ has i=3 (it dissociates in Ca⁺ and 2 Cl⁻). The lower concentration is in the left side so the osmotic flow proceeds from left to right.

0.010 M CaCl2 (aq) | 0.010 M FeCl3 (aq) : CaCl₂ has i=3 and FaCl₃ has i=4 (it dissociates in Fe⁺ and 3 Cl⁻), so the lower concentration is in the left side and osmotic flow proceeds from left to right.