18g glucose is dissolved in 180g of water. The relative lowering of vapour pressure is ?

a)1 b)1.8
c)0.01 d)0.001

Question

18g glucose is dissolved in 180g of water. The relative lowering of vapour pressure is ?

a)1  b)1.8
c)0.01  d)0.001

Henry Antrim · Accepted Answer

See Below

Vapor Pressure lowering is governed by Raoult's law, which basically states: #P_n = (X_n)xxP_o# were #P_n# is the new vapor pressure, #X_n# is the mole faction of the solvent, and #P_o# is the pure solvent vapor pressure. I think the relative lowering would be#(1 - P_n/P_o)#, which is #(1-X_n)# #X_n = ("moles of water")/("total moles")# #X_n = 10/10.1 = 0.99009901# Relative Lowering: #(1-0.99009901) = 0.00990099 = 0.01#

Jeremiah Adams · Answer

undefined The relative lowering of vapor pressure can be calculated using the formula:

$$ \text{Relative lowering of vapor pressure} = \frac{\text{Moles of solute}}{\text{Moles of solvent}} $$

First, we need to find the number of moles of glucose (solute) and water (solvent) in the solution.

Given that the molar mass of glucose (C6H12O6) is approximately 180 g/mol, we have:

$$ \text{Number of moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ moles} $$

Since water is the solvent, its molar mass is approximately 18 g/mol. Therefore:

$$ \text{Number of moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} $$

Now, using the formula for relative lowering of vapor pressure:

$$ \text{Relative lowering of vapor pressure} = \frac{\text{Moles of solute}}{\text{Moles of solvent}} = \frac{0.1}{10} = 0.01 $$

So, the correct answer is option c) 0.01.

18g glucose is dissolved in 180g of water. The relative lowering of vapour pressure is ? a)1 b)1.8 c)0.01 d)0.001