A semipermeable membrane (SPM) contains an aqueous solution of 0.07 M sucrose, 0

Question

A semipermeable membrane (SPM) contains an aqueous solution of 0.07 M sucrose, 0.01 M glucose, and 0.1 M NaCI. It is immersed in a beaker containing a different aqueous solution [0.05 M sucrose, 0.03 M glucose, 0.1 M NaCI). The SPM allows water and glucose and NaCI to cross, but is impermeable to the larger disaccharide sucrose. Circle all that apply. 1a. Which of the original solutions is isotonic with a normal cell? SPM or beaker 2. Will NaCI cross the SPM? Yes or No 3. Which solute(s) will exhibit a net diffusion into the cell? Sucrose Glucose NaCI 4. Over time, the volume of the bag will increase decrease stay the same Calculate the following. You MUST show your work. Lidocaine is a topical anesthetic, C_14H_22N_2O (234g/mol) a. Number of moles in 0.3 grams of lidocaine. b. Number of grams of lidocaine in 0.2 moles c. Lidocaine is administered as a 1% (w/w) solution. How many grams of lidocaine are in a liter of this solution. Density is 1.0 g/mL. d. Number of molecules of lidocaine given to patient if they receive a 1 mL injection of 1% lidocaine solution. Given the following correctly BALANCED reaction: C_7H_16 + 11 O_2 rightarrow 7CO_2 + 8H_2O delta H = 1078 Kcals/mol How grams of carbon dioxide are produced from 600 grams of C_7H_16 (heptane). How many moles of heptane will heat 100 grams of water from 20 degree C to 80 degree C? When 900 grams of heptane (C_7H_16) are mixed with 100 moles of oxygen, which substance is the limiting reagent? ___________ How many grams of the excess reagent remain after complete reaction? _______________

Explanation / Answer

1) both the solution will be isotonic as number of molecules is same in both the solutions

No, NaCl will not cross

2) Moles = Mass / Molecular weight

Moles = 0.3 / 234 = 0.00128 moles

b) mass = Moles X molecular weight = 0.2 X 234 = 46.8 grams

c) 1% w /w means 1gram in 100 grams of solution

1L = Mass / Density

Mass = 1000mL X 1 = 1000grams

so mass of solute = 10grams in 1 litre

d) MAss of lidocaine = 10grams in 1000 mL

so mass in 1mL = 10 X 10^-3 grams

Moles = Mass / Molecular weight = 0.01 / 234 = 4.27 X 10^-5 moles

Number of molecules in one mole = 6.023 X 10^23

Number of molecules in 4.27 X 10^-5 moles = 4.27 X 10^-5 X 6.023 X 10^23 = 2.571 X 10^19 molecules

6) As per the given equation 1mole of C7H16 will give 7moles of CO2

Molecular weight of C7H16 = 100g / mole

So 600 grams = 6 moles of C7H16, which means the mole of CO2 produced = 6 X 7 moles = 6 X 7 X 44grams

mass of CO2 produced = 1848 grams

b) the heat required = Mass of water X specific heat X change in temperature

Heat required = 100 x 4.186 X 60 = 25116 Joules

Moles of heptane required = Heat required / enthalpy of reaction of heptane = 25116 / 1078 = 23.298 moles

8) as per given equation 1mole of heptane will react with 11 moles of O2

900grams of heptane = 9 moles of heptane

it will react with = 11 X 9 = 99 moles of O2

So the limiting reagent is heptane as O2 is present in excess

9) the excess reagent left = 1mole of O2 = 32 grams of O2

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