Homework Answers
given that 150 mM of NaCl i.e., 150 m moles/L
so it required 150 milli moles to prepare 1 L of the solution
1 mole of NaCl has 58.5 g of NaCl
150 milli moles of NaCl has 58.5g * 150 * 10^-3 mole/1mole
mass of NaCl = 8.775 g
Osmolarity is defined as the number moles of solute particles dissolved in given solution.
NaCl -----> Na+ + Cl-
Osmolarity for the given solution = 2 * 150 m mol/L = 300 mmol/L
Osmolarity of the solution = 0.3 mol/L
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(17) Suppose you have stock solution with a solute concentration of 10 mm. If you wanted to dilute this solution to make a total of 1 ml of a 3 mM solution, what volume of stock and what volume of water would you use? Give your answer in microliters (ul), and show your calculations. Volume of stock3 ul Volume of water · T ul
Please help!!!
Drive 9 47 PM Mon Jan 21 30% Case Study Questions Dehydration is a serious and common condition. It changes body fluid volumes and can change body fluid composition and concentration. This can result in the redistribution of fluid between compartments. This problem deals with osmosis between red blood cells and ECF A patient was admitted to the ER late at night. The residents were asleep leaving a 3nd year medical student (Jim Stephens) on duty. The patient...
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equilibrium constant? Please provide explanation for steps.
Thank you!
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