Question

the mass of urea that would be dissolved in 180 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water is

• 10 g
• 5 g
• 3.5 g
• 6 g​

Answer 1

Answer:

The mass of urea that would be dissolved in 180 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water is • 10 g. 5 g.

Answer 2

The mass of urea that would be dissolved in 180 g of water to produce the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water is 62g.

Given:

Mass urea in 180g of water produces the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water.

To Find:

The mass of urea would be dissolved in 180 g of water to produce the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water.

Solution:

To find the mass of urea we will follow the following steps:

As we know,

Lowering in vapour pressure = mole fraction of solute.

$\frac{p - p0}{p} = x2$

p is the final pressure of the solution and p0 is the initial pressure, x2 is the mole fraction of solute.

$x2 = \frac{n2}{n1 + n2}$

n1 and n2 are several moles of solvent and solute respectively.

The molecular mass of urea = 60g

number of moles of urea =

$\frac{w}{60}$

Here, w is the weight of urea added.

The molecular mass of cane sugar = 342g

number of moles of cane sugar =

$\frac{90}{342} = 0.26$

Moles of water in urea =

$\frac{180}{18} = 10$

Moles of water in cane sugar =

$\frac{100}{18} = 5.55$

Mole fraction of urea =

$\frac{ \frac{w}{60} }{ \frac{w}{60} + 10}$

Mole fraction of cane sugar =

$\frac{0.26}{0.26 + 5.55} = \frac{0.26}{5.81} = 0.044$

As given lowering in vapour pressure of urea and sugar is the same, so equating the mole fraction of urea and sugar cane we get,

$\frac{ \frac{w}{60} }{ \frac{w}{60} + 10} = 0.044$

$\frac{w}{60} = 0.044 \times ({ \frac{w}{60} + 10})$

$\frac{w}{60} = \frac{0.044w + 60}{60}$

$w - 0.044w = 60$

$0.956w = 60$

$w = \frac{60}{0.956} = 62g$

Henceforth, the mass of urea that would be dissolved in 180 g of water to produce the same lowering of vapour pressure as is produced by dissolving 90 g of cane sugar ( C12H22O11) in 100 g of water is 62gram.