find the concentration of glucose that is equivalent 6gm/l urea solution ...molecular mass of glucose 180 and urea 60
Answers
Answer: π = CRT where π is the osmotic pressure of the solution C = Molarity of ... two solutions to be isotonic, their osmotic pressures have to be equal. π1 = π2 ... and solving for M we get, M = 300 g/mol Hence, the correct option is B. ... A solution containing 10 g/dm3 of urea (molecular mass = 60 g ... M=300g/mo
Explanation:
Given:
The concentration of urea = 6 gm/L
M.wt of glucose = 180 gm
M.wt of urea = 60 gm
To Find:
The concentration of glucose equivalent to the given concentration of urea.
Calculation:
- The Molarity of urea, M1 = (6 × 1000) / (60 × 1000)
⇒ M1 = 1 / 10 = 0.1 M
- Now applying this condition to glucose, we get:
M1 = (w × 1000) / (180 × 1000)
⇒ 0.1 = w / 180
⇒ w = 18 gm
⇒ 1 litre of glucose solution contains 18 gm glucose.
- So, the concentration of glucose that is equivalent to the given concentration of urea is 18 gm/L