At a certain temperature 5% w/v solution of glucose (molar mass = 180) is isotonic with 8% W/v solution of an unknown solute. What is the molar mass of the unknown solute?
a. 36
b. 150
c. 216
d. 288
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Given:
Mass/volume % of glucose = 5 % w/v
Molar mass of glucose = 180 gm/mol
Mass/volume % of unknown solution = 8 % w/v
This solution is isotonic with the solution of an unknown solute.
To Find:
The molar mass of the unknown solute.
Calculation:
- Molarity of Glucose = (5 × 1000)/(180×100) = 5/18
- Molarity of unknown solute = (8 × 1000)/(M.wt × 100) = 80/M.wt
- Isotonic solutions have the same osmotic pressure.
⇒ Π (glucose) = Π (unknown)
⇒ C1RT = C2RT
⇒ 5/18 = 80/M.wt
⇒ M.wt = (80 × 18)/5
⇒ M.wt = 288 gm
- So, the correct answer is option (d) 288 gm.
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