At 27° C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is g mol -1 (Nearest integer) (Given: R =0.083 L bar K -1 mol -1 )

Question

At 27° C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is g mol -1 (Nearest integer) (Given: R =0.083 L bar K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

= CRT (400 Pa /105)=((2.5 g / M o )/250 / 1000 L ) × 0.83 ( L - bar / K ⋅ mol ) × 300 K M o =62250

Q. At $2 7^{\circ} C$ , a solution containing $2.5 g$ of solute in $250.0 m L$ of solution exerts an osmotic pressure of $400 P a$ . The molar mass of the solute is ____ $g m o l^{- 1}$ (Nearest integer)
(Given : $R = 0.083 L ba r K^{- 1} m o l^{- 1}$ )

$= CRT$
$\frac{400 P a}{1 0 ^{5}} = \frac{\frac{2.5 g}{M _{o}}}{250/1000 L} \times 0.83 \frac{L - ba r}{K \cdot m o l} \times 300 K$
$M_{o} = 62250$

Q. At 27∘C, a solution containing 2.5g of solute in 250.0mL of solution exerts an osmotic pressure of 400Pa. The molar mass of the solute is ____gmol−1 (Nearest integer) (Given : R=0.083LbarK−1mol−1 )

=CRT 105400Pa​=250/1000LMo​2.5g​​×0.83K⋅molL−bar​×300K Mo​=62250

Q. At $2 7^{\circ} C$ , a solution containing $2.5 g$ of solute in $250.0 m L$ of solution exerts an osmotic pressure of $400 P a$ . The molar mass of the solute is ____ $g m o l^{- 1}$ (Nearest integer)
(Given : $R = 0.083 L ba r K^{- 1} m o l^{- 1}$ )

$= CRT$
$\frac{400 P a}{1 0 ^{5}} = \frac{\frac{2.5 g}{M _{o}}}{250/1000 L} \times 0.83 \frac{L - ba r}{K \cdot m o l} \times 300 K$
$M_{o} = 62250$