Question

the vapor pressure of pure water at 50°c is 92.5 mm Hg. a solution containing sucrose has a vapor pressure of 90.8 mm Hg at 50°c. a. What is the boiling point elevation of this solution? b. if it has density of 1.08 g/mL, what is the osmotic pressure of this solution at 50°c?​

Answer 1

Answer:

According to Raoult's law,

Po−Ps/Po = n/n+N

or △p= n/n+N.p0

Given: n= 50/342 = 0.146;N = 500/18 =27.78

and Po = 23.8

Substituting the value in the above equation,

△p= 0.146/0.146+27.78 × 23.8=0.124 mmHg

Explanation:

Hope its helpful to you .

Answer 2

The osmotic pressure of the given solution at 50°C is .

Given:

The vapor pressure of pure water at 50°C = 92.5 mm Hg.

The vapor pressure of sucrose solution at 50°C = 90.8 mm Hg.

The density of sucrose solution = 1.08 g/mL.

The colligative constant ($K_b$) of pure water = 0.520°C/m.

The molar mass of pure water = 18.0 g/mol.

The molar mass of sucrose solution = 342 g/mol.

To Find:

We have to find the following:

1. The boiling point elevation of the sucrose solution.

2. The osmotic pressure of the sucrose solution at 50°C.

Solution:

1. The boiling point elevation of the sucrose solution.

The mole fraction of water, $X_{Water} = \frac{P_{Sucrose } }{P^0_{Water} }$

Substituting the given values, we get,

$X_{Water} = \frac{90.8 }{92.5 } = 0.982$.

The number of moles of water can be found as,

$n_{Water} = X_{Water}$ × $Molar mass of water = 0.982$ × $18 = 17.7g.$

Similarly, the number of moles of sucrose solution is given by,

$n_{Sucrose} = 0.018 mol.$

The molality (m) of the sucrose solution can be calculated using the equation,

$m = \frac{0.018}{0.0177} = 1.02m.$

∴, The expression for the boiling point elevation of the sucrose solution, Δ$T_b = i. K_b.m$,

where, $i$ is the number of moles of the particles in the solution, $K_b$ is the colligative constant and $m$ is the molality.

On substituting the given values in the above equation, we get,

Δ$T_b = 1$ × $0.520$ × $1.02 = 0.530$ °C.

2. The osmotic pressure of the sucrose solution at 50°C.

From above, $X_{Water} = 0.982$ and $n_{Water} = 17.7g$.

$n_{Sucrose} = 0.018$ × $342 = 6.2 g$.

On adding the values of $n_{Water}$ and $n_{Sucrose}$, we get,

$n_{Water} + n_{Sucrose} = 6.2 + 17.7 = 23.9 g$

Hence, the osmotic pressure of the given solution at 50°C is .

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