Question

A solution is prepared by dissolving 10g naoh in 1250ml of solvent of density 0.8 g/mol the molality of the solution is?

Answer 1

Answer:

0.25 MOLAL.

Explanation:

SEE YOU HAVE 10 GRAM NaOH.

AND MOLAR MASS OF IT IS = 40 GRAM.

SO NO. OF MOLES= 10/40 = 1/4 MOLE.

NOW VOLUME OF SOLVENT= 1250ML AND DENSITY= 0.8 G/ML.

SO MASS OF SOLVENT= 0.8 X 1250= 1000 GRAM= 1 KG.

SO MOLALITY= MOLE OF SOLUTE/ MASS OF SOLVENT IN KG.

=1/4/1= 1/4 MOLAL= 0.25 MOLAL.

Answer 2

Ans;;

755 m

Explanation:

molality =  wt of solute(g) / wt of solvent (g) *1000/molar mass of solute

density of solvent= mass / volume

0.8=mass/1250

mass=1000 g

molality = 10/1000*1000/75.5

since the molar mass of nacl is 40+35.5=75.5

therefore molality is 755 m