Calculate the molar mass of a solute when $4 \ g$ of it is dissolved in $1 \ dm^3$ of solvent,having an osmotic pressure of $2 \ atm$ at $300 \ K$. $[R = 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}]$
- A$49.2 \ g \ mol^{-1}$
- B$44.5 \ g \ mol^{-1}$
- C$54.2 \ g \ mol^{-1}$
- D$56.4 \ g \ mol^{-1}$
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When does the osmotic pressure of a solution increase?
Easy
View SolutionWhat is the osmotic pressure of a semi-molar solution at $27^{\circ} C$ (in $atm$)?
$(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
Isotonic solutions have the same
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View SolutionCalculate the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 \ g$ of polymer of molar mass $185,000$ in $450 \ mL$ of water at $37^{\circ} C$.
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View SolutionAt $27^{\circ} C$,the osmotic pressure of a solution containing $4 \ g$ of a non-electrolyte solute in $1.0 \ L$ of solution is $0.4 \ bar$. The molar mass of the solute in $g \ mol^{-1}$ is :
$(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1})$
$(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1})$
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