100 mL of a solution containing 0.4 g urea ( NH 2 CONH 2) and 0.9 g of another substance ( X ) shows an osmotic pressure of 2.85 atm at 25° C. Molar mass of the substance ( X ) is

Question

100 mL of a solution containing 0.4 g urea ( NH 2 CONH 2) and 0.9 g of another substance ( X ) shows an osmotic pressure of 2.85 atm at 25° C. Molar mass of the substance ( X ) is (A) 85 g mol -1 (B) 92 g mol -1 (C) 180 g mol -1 (D) 140 g mol -1

Tardigrade Admin · Accepted Answer

Osmotic pressure, π=(M1+M2) R T M1 ( urea ) =(0,4 g/60 g mol -1 × 0.1 L ) =(1/15) mol L -1 M2(X)=(0.9 g /m g mol -1 × 0.1 L ) =(8/m) mol L -1 π=2.85 atm (given) ∴ 2.85=((1/15)+(9/m)) × 0.0821 × 298 ((1/15)+(9/m))=0.1165 (9/m)=(0.1165-(1/20)) (9/m)=0.0498 ∴ m =180 g mol -1

Q. $100 m L$ of a solution containing $0.4 g$ urea $(N H_{2} CON H_{2})$ and $0.9 g$ of another substance $(X)$ shows an osmotic pressure of $2.85$ atm at $2 5^{\circ} C$ . Molar mass of the substance $(X)$ is

$85 g m o l^{- 1}$

$92 g m o l^{- 1}$

$180 g m o l^{- 1}$

$140 g m o l^{- 1}$

Q. 100mL of a solution containing 0.4g urea (NH2​CONH2​) and 0.9g of another substance (X) shows an osmotic pressure of 2.85 atm at 25∘C. Molar mass of the substance (X) is

85gmol−1

92gmol−1

180gmol−1

140gmol−1

Osmotic pressure, π=(M1​+M2​)RT M1​ ( urea ) =60gmol−1×0.1L0,4g​ =151​molL−1 M2​(X)=mgmol−1×0.1L0.9g​ =m8​molL−1 π=2.85 atm (given) ∴2.85=(151​+m9​)×0.0821×298 (151​+m9​)=0.1165 m9​=(0.1165−201​) m9​=0.0498 ∴m=180gmol−1

Q. $100 m L$ of a solution containing $0.4 g$ urea $(N H_{2} CON H_{2})$ and $0.9 g$ of another substance $(X)$ shows an osmotic pressure of $2.85$ atm at $2 5^{\circ} C$ . Molar mass of the substance $(X)$ is

$85 g m o l^{- 1}$

$92 g m o l^{- 1}$

$180 g m o l^{- 1}$

$140 g m o l^{- 1}$