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One mole of sodium chloride is added to 100 ml of water in beaker A. One mole of

ID: 1055365 • Letter: O

Question

One mole of sodium chloride is added to 100 ml of water in beaker A. One mole of glucose, C_6H_12O_6, is added to 100ml of water in beaker B. How much more will the freezing point of water be lowered in beaker A than in beaker B? Two thirds as much twice as much three times as much Beakers A and B will experience the same degree of freezing point depression. It is impossible to determine. Which of the following correctly describes what happens when red blood cells are placed in a hypertonic solution? There will be a net movement of water into the red blood cells, the cells will swell and collapse. There will be a net movement of water from the hypertonic solution to the red blood cells, and the cells will shirr and burst. There will be no net movement of water because red blood cells donor have a semipermeable membrane. There will be no net movement of water because the red blood cells and the hypertonic solution have the same osmolarity. Na_3PO_4 dissolves in water to produce an electrolytic solution. What is the osmolarity of a 2.0 times 10^-3 M Na_3PO_4 solution? 1.0 times 10^-3 osmolar 2.0 times 10^-3 osmolar 4.0 times 10^-3 osmolar 5.0 times 10^-3 osmolar 8.0 times 10^-3 osmolar Increasing the pressure of a gas over a solution increases the solubility of the gas in the solution. This is an example corresponding to which law? Dalton's law Henry's law Tyndall's Law Raoult's Law Boyle's law Assuming that air is a solution containing four molecules of N_2 for every one molecule O_2, what is the concentration of O_2 in this solution, expressed as % (m/m)[Formula weights: N_2, 28.02 amu; 32.00 amu] 22%(m/m) 47%(m/m) 53% (m/m) 78%(m/m) 114%(m/m) Which of the following is Not a colligative property of a solution? Vapor pressure lowering. Solubility boiling point elevation freezing point depression osmotic pressure What concentration term is defined as the number of moles of solute per kilogram of solvent in a solution? Normality osmolarity %(m/v) molarity molality

Explanation / Answer

48- depression in freezing point is a colligative property which depends on only number of solute particles and not on nature of solute particles in a solution. Here NaCl break into Na+ and Cl- in solution, do effectively each one mole of NaCl gives 2 mol of particles in solution(vant hoff factor=2).

So twice more depression in case of NaCl becoz it gives 2 mol of particles whereas sugar gives only one mole in solution as it doesnot break

Hence option- B

49- B option is correct because in case of hypertonic solution more water is gven out.

50- 1 mol Na3PO4 gives 4 mol of particles, so 2 millimole would give 8 millimole of particles. Hence E is correct option.

51- B

52- N2:O2 = 4:1 when comparing their number of moles.

In terms of mass, this ratio is (4×28)/32

Hence mass% of O2 =(32×100)/(4×28 + 32) ~22%

53- B

54- E

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