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An aqueous solution of NaCI freezes at -...

An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:

A

0.0585 K

B

0.0512 K

C

1.864 K

D

0.0265 K

Text Solution

Verified by Experts

The correct Answer is:
B
`(DeltaT_f)/(DeltaT_b)=K_f/K_b DeltaT_b=(K_bxxDeltaT_f)/K_f`
`DeltaT_b=(0.512xx0.186)/1.86=0.0512`
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