Home
Class 12
CHEMISTRY
58.5 g of NaCl and 180 g of glucose were...

58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (bp) of the resulting solutions.

A

NaCl solution will show higher elevation of b.pt.

B

Glucose solution will show higher elevation of b.pt.

C

Both the solutions will show equal elevation of b.pt.

D

The b.pt. of elevation will be shown by neither of the solutions

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaT_(b)=iK_(b)m`
For water implies 1000 mL = 1000g
Molality of NaCl = `(w//M.W.)/(W("solvent"))xx1000`
`=(58.5//58.5)/(1000)xx1000=1m`
Molality of glucose `=(180//180)/(1000)xx1000=1m`
i for NaCl = 2, i for glucose = 1
`DeltaT_(b)` for `NaClgtDeltaT_(b)` for glucose.
Promotional Banner

Similar Questions

Explore conceptually related problems

The molarity of a solution having 18g of glucose dissolved in 500 g of water is

5.85 g NaCl is dissolved in 500 ml of water. The molarity is

5.85 g of NaCl are dissolved in 90 g of water. The mole fraction of NaCl is-

If 23 g of glucose ( molecular mass 180)is dissolved in 60ml of water at 15^(@)C , then the osmotic pressure of this solution will be

1.80 g of glucose (molar mas =180) is dissolved in 36.0 g of water in a beaker. The total number of oxygen atoms in the solution is