Question

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​

Answer 1

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

Answer 2

Answer:

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