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5 1 The answers are A, B, and C These waste products are all normally excreted in large amounts; a decreased GFR would cause their plasma concentrations to increase until the filtered load was increased enough to reestablish normal excretion In contrast, the reabsorption Tms for glucose, amino acids, and many other organic compounds that are not waste products are usually so high relative to normal filtered loads that, even with a 50% loss of nephrons, virtually all the filtered loads are reabsorbed Accordingly, their plasma concentrations are virtually independent of renal function (ie, the kidneys do not participate in the setting of their plasma concentrations) The answer is no The overall tubular handling of urea is reabsorption Urinary urea concentration is higher than that of plasma because relatively more water has been reabsorbed than urea, thereby concentrating the urea in the tubular fluid leaving the kidney This is simply the GFR (180 L/day) times the concentration in the filtrate: 180 L/day 3 0005 g/dL 3 10 dL/L 5 9 g/day The answer is A The luminal membranes of nephron segments beyond the proximal tubule do not express glucose transporters, so no further reabsorption occurs regardless of conditions Decreased pH (acidify the urine) Weak bases like quinine become protonated, and thus less permeable, at low pH This would prevent its passive reabsorption and increase its excretion (a) About 50%; (b) about 100% (due to secretion in the thin limbs); (c) about 50%

E911

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+ (7000 mL)(20 mol K3PO4/l000 mL)(3 mol K / l mol K3PO4)

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POTS & VOP VOP POTS TR-008/ GR-303 Nextgen DLC DS-1, DS-3, OC-3

The total potassium is 450 mol in a total volume of 1000 L Thus, the potassium concentration is 450 M

6 1 6 2 116 g/day The answer is B Mannitol is freely filtered and neither secreted nor reabsorbed It acts as an excess osmole in the same way that glucose does when its filtered load exceeds the Tm for reabsorption Sodium excretion will increase and for the same reasons as when there is excess glucose

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3 C The net ionic equation is: Sr 2+ ( aq) + SO2 ( aq) SrSO 4 ( s) 4 The strontium nitrate solution contains: (700 mL)(020 mol Sr(NO3)2/1000 2+ (1 mol Sr /1 mol Sr(NO3)2) = 0014 mol Sr

Answers to study Questions / 205

POTS & VOP VOP POTS TR-008/ GR-303

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7 A To calculate the molality of a solution, both the moles of solute and the kilograms of solvent are needed A liter of solution would contain a known number of moles of solute To convert this liter to mass, a mass to volume relationship (density) is needed 8 E The mole fraction may be determined by dividing the vapor pressure of the desired substance by the sum of all the vapor pressures 87 mm Hg/(87 mm Hg + 170 mm Hg) = 034 9 C To produce a molar solution of any type, the final volume must be the desired volume This eliminates answers D and E B involves mass of water instead of volume A calculation of the required mass will allow a decision between A and B (30 L)(020 mol K3PO4/L)(212 g K3PO4/1 mol K3PO4) = 130 g K3PO4 10 C A 52 molal solution has 52 mol of methyl alcohol in one Kg (1000 g) of water The moles of water are needed (1000 g H2O)(1 mol H2O/180 g H2O) = 556 mol H2O The mole fraction may now be determined (52 mol CH3OH)/(52 + 556) mol = 0086 11 B The boiling point depends on the boilingpoint elevation (a colligative property) All colligative properties depend on the concentration of particles present C is a nonelectrolyte, thus, the concentration of the particles is 020 M All of the other substances are strong electrolytes The concentration of particles in each of these may be determined by multiplying the molarity given by the van t Hoff factor Each of these is calculated as follows: A: 2 010 = 020 D: 2 010 = 020 12 D (50 mole MgSO4/1000 mL)(1000 mL) (1204 g MgSO4/mol MgSO4) = 60 g MgSO4 B: 4 010 = 040 E: 2 010 = 020.

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