Question

What is the molarity (m) of a solution prepared by dissolving 29.25g NaCl in 3.00kg of water?

a. 0.0002
b. 0.17
c. 1.53
d. 9.75

Answer 1

number of moles of $NaCl=\frac{29.25}{58.5}=0.5mol$

mass of solvent(water)$=3Kg$

molality, $m=\frac{moles of solute}{mass of solvent(in Kg)}=\frac{0.5}{3}=0.17molal$