calculate the molality of a 3 molar aquous solution of NaCl whose density is 0.9g m /L...dont copy paste google answer i already searched
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Explanation:
Molarity =3M,
Density = 0.9g/L
=9 ×10^(-5)g/ml
Formula Mass of NaCl =58.5u
Now,
molality= M×1000/d×1000-M×Formula Mass
So, m =3000/9×10^(-2) - 175.5
m= 3000/17541×10^(-2)
m=300000/17541
m= 17.1m
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