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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 (b.pt.) 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

`Delta Tb = i Kbm`
For water `rArr 1000 mL = 1000 g`
Molality of `NaCl = (w//M.W)/(W(Solvent) xx 1000`
`= (58.5 // 58.5)/(1000) xx 1000 =1m`
Molality of gulcose `=(180//180)/(1000) xx 1000 =1m`
if For `NaCl =2` , i for glucose = 1
`Delta T_b` for `NaCl gt Delta T_b` for glucose
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