Question

If the sea water is 3.5% (w/w) aqueous solution
of NaCl then its molality will be :-
(1) 0.598
(2) 36.27 m
(3) 0.62 m
(4) 0.578 m​

Answer 1

Answer:

3.5%w/w .means 3.5 grams of Nacl present in 100 grams of solution

then mass of solvent is 100-3.5=96.5grams

molality =no.of moles of Nacl/mass of solvent

no.of moles =3.5/58.5

after solving

option (3)