Question

A solution has an osmotic pressure of 'x' kPa at 300 K having 1 mole of solute in 10.5 m³ of solution. If it's osmotic pressure is reduced to `(1/10)^"th"` of it's initial value, what is the new volume of solution?

पर्याय

• 11.0 m³

• 105 m³

• 30 m³

• 110 m³

Answer 1

105 m³

Explanation:

The given values are

π₁ = 'x' kPa, π₂ = `"x"/10` kPa, T₁ = 300 K, n₁ = 1 mole

Using formula,

π₁V = x₁RT

x × 10.5 = 1 × 8.314 × 300

x = `(8.314 xx 300)/10.5`

If osmotic pressure is reduced `(1/10)^"th"` then, value π₂ = `"x"/10` kPa

π₂V = nRT

`"x"/10 xx "V"` = n × R × T

`(8.314 xx 300)/(10.5 xx 10) xx "V"` = 1 × 8.314 × 300

V = 10.5 × 10

V = 105 m³