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A 0.6% solution of urea (molecular weight = 60) would be isotonic with
- a)0.1 M glucose
- b)0.1 M KCl
- c)0.6% glucose solution
- d)0.6% KCl solution
Correct answer is option 'A'. Can you explain this answer?
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A 0.6% solution of urea (molecular weight = 60) would be isotonic with...
Explanation:
Isotonic solutions are those that have the same osmotic pressure, i.e., the same concentration of solute particles, as the solution to which they are being compared. In this case, we are looking for a solution that is isotonic with a 0.6% solution of urea.
Step 1: Calculate the molarity of 0.6% urea solution
0.6% solution means 0.6 g of urea in 100 mL of solution.
Molecular weight of urea = 60 g/mol
Number of moles of urea in 100 mL of solution = (0.6/60) = 0.01 mol
Molarity of 0.6% urea solution = 0.01/0.1 = 0.1 M
Step 2: Determine the solute concentration of each option
a) 0.1 M glucose
Molecular weight of glucose = 180 g/mol
0.1 M solution means 0.1 moles of glucose in 1 liter of solution
Mass of glucose in 1 liter of solution = (0.1 mol) x (180 g/mol) = 18 g/L
b) 0.1 M KCl
Molecular weight of KCl = 74.5 g/mol
0.1 M solution means 0.1 moles of KCl in 1 liter of solution
Mass of KCl in 1 liter of solution = (0.1 mol) x (74.5 g/mol) = 7.45 g/L
c) 0.6% glucose solution
0.6% solution means 0.6 g of glucose in 100 mL of solution
Mass of glucose in 1 liter of solution = (0.6 g) x (10) = 6 g/L
d) 0.6% KCl solution
0.6% solution means 0.6 g of KCl in 100 mL of solution
Mass of KCl in 1 liter of solution = (0.6 g) x (10) = 6 g/L
Step 3: Compare the solute concentration of each option with 0.1 M urea solution
The solute concentration of option 'a' (18 g/L) is closest to that of 0.1 M urea solution (6 g/L) and therefore, it is the correct answer. The solute concentration of options 'c' and 'd' is too low, while the solute concentration of option 'b' is too high to be isotonic with 0.1 M urea solution.
Therefore, option 'A' is the correct answer.
Isotonic solutions are those that have the same osmotic pressure, i.e., the same concentration of solute particles, as the solution to which they are being compared. In this case, we are looking for a solution that is isotonic with a 0.6% solution of urea.
Step 1: Calculate the molarity of 0.6% urea solution
0.6% solution means 0.6 g of urea in 100 mL of solution.
Molecular weight of urea = 60 g/mol
Number of moles of urea in 100 mL of solution = (0.6/60) = 0.01 mol
Molarity of 0.6% urea solution = 0.01/0.1 = 0.1 M
Step 2: Determine the solute concentration of each option
a) 0.1 M glucose
Molecular weight of glucose = 180 g/mol
0.1 M solution means 0.1 moles of glucose in 1 liter of solution
Mass of glucose in 1 liter of solution = (0.1 mol) x (180 g/mol) = 18 g/L
b) 0.1 M KCl
Molecular weight of KCl = 74.5 g/mol
0.1 M solution means 0.1 moles of KCl in 1 liter of solution
Mass of KCl in 1 liter of solution = (0.1 mol) x (74.5 g/mol) = 7.45 g/L
c) 0.6% glucose solution
0.6% solution means 0.6 g of glucose in 100 mL of solution
Mass of glucose in 1 liter of solution = (0.6 g) x (10) = 6 g/L
d) 0.6% KCl solution
0.6% solution means 0.6 g of KCl in 100 mL of solution
Mass of KCl in 1 liter of solution = (0.6 g) x (10) = 6 g/L
Step 3: Compare the solute concentration of each option with 0.1 M urea solution
The solute concentration of option 'a' (18 g/L) is closest to that of 0.1 M urea solution (6 g/L) and therefore, it is the correct answer. The solute concentration of options 'c' and 'd' is too low, while the solute concentration of option 'b' is too high to be isotonic with 0.1 M urea solution.
Therefore, option 'A' is the correct answer.
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A 0.6% solution of urea (molecular weight = 60) would be isotonic witha)0.1 M glucoseb)0.1 M KClc)0.6% glucose solutiond)0.6% KCl solutionCorrect answer is option 'A'. Can you explain this answer?
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