Question

34.2 g of cane sugar is dissolved in 180 g of water.

The relative lowering of vapour pressure will be​

Answer 1

Explanation:

Given 34.2 g of cane sugar is dissolved in 180 g of water.  The relative lowering of vapour pressure will be​

• Molar masses of cane sugar and water are 342 g / mol and 18 g / mol respectively.
• Therefore number of moles of cane sugar will be 34.2 g / 342 g/mol = 0.1 mol
• Number of moles of water = 180 g / 18 g / mol = 10 mol
• Therefore mole fraction of cane sugar will be 0.1 / 0.1 + 10 = 0.0099
• Hence the relative lowering of vapor pressure of solution is equal to the mole fraction of cane sugar.

Therefore relative lowering of vapour pressure will be 0.0099

Reference link will be

https://brainly.in/question/3933993

Answer 2

The relative lowering of vapor pressure is 0.0099

Explanation:

The molar mass of the cane sugar is 342 g/mol

Number of moles of cane sugar = 34.2/342 = 0.1 moles

The molar mass of the water is 18 g/mol

Number of moles of water = 180/18 = 10 moles

The relative lowering of vapor pressure is equal to mole fraction of cane sugar.

Now,

Mole fraction of cane sugar = 0.1/(0.1 + 10) = 0.0099

Thus, the relative lowering of vapor pressure is 0.0099