If 3 g of glucose (molar mass =180 g ) is dissolved in 60 g of water at 15° C, the osmotic pressure of the solution will be

Question

If 3 g of glucose (molar mass =180 g ) is dissolved in 60 g of water at 15° C, the osmotic pressure of the solution will be (A) 6.57 atm (B) 5.57 atm (C) 0.34 atm (D) 0.65 atm

Tardigrade Admin · Accepted Answer

π= C . R . T =( w 2/ M 2) (1000/ V ( m ℓ)) × R . T ⇒ (3/180) × (1000/60) × 0.0821 × 288=6.568 atm

Q. If $3 g$ of glucose (molar mass $= 180 g)$ is dissolved in $60 g$ of water at $1 5^{\circ} C$ , the osmotic pressure of the solution will be

6.57 atm

5.57 atm

0.34 atm

0.65 atm

$π = C . R . T = \frac{w _{2}}{M _{2}} \frac{1000}{V ( m ℓ )} \times R . T$
$\Rightarrow \frac{3}{180} \times \frac{1000}{60} \times 0.0821 \times 288 = 6.568 a t m$

Q. If 3g of glucose (molar mass =180g) is dissolved in 60g of water at 15∘C, the osmotic pressure of the solution will be