Molarity and molality of pure ch3cooh are respectively

Question

Please log in or register to answer this question.

Parulgupta · Answer 1 · 2023-09-15T04:43:27+0000

\(M=\frac{n}{v}\)

M = molar concentration

n = moles of solute

v = liters of solution

\(M=\frac{1}{\frac{60}{1.5}}\times100\)

\(m=\frac{mol}{kg}\)

m = molality

mol = moles of solute

{kg} = kilogram of solvent

\(m=\frac{1}{60}\times1000\)

= 16.67