Answer to Question #345823 in General Chemistry for batman

Question

Calculate the molality of a 0.410 molar aqueous solution whose density is 1.15 g/mL.

Answer 1

I think the question is as follows (comment:

if there is a solution of another substance, the calculation should be made by substituting the molecular mass of that substance for 180) :

Calculate the molality of a 0,410 molar aqueous glucose solution whose density

is 1,15 g/mol.

1.

the volume of the solution is not given. so we consider it as 1 liter and find the mass of glucose in it:

n — amount of substance (mol);

C_M — concentration of molarity (mol/l);

V — volume of solution (l);

n= C_M * V =0,41mol/l * 1 l =0,41mol glucose

m = n * M_r (C₆H₁₂O₆) = 0,41 * 180 = 73,8 gr

0,41 mol (73,8 gr) glucose

2.

To find the concentration of molality, we divide the amount of solute (mol) by the mass (kg) of the solvent.

a)

for this we first determine the mass of the solvent. that is, we separate the mass of glucose from the mass of the solution:

Mass of solution = p * V (ml) = 1,15g/ml * 1000ml = 1150 gr

Mass of solvent = 1150 – 73,8 = 1076,2gr = 1,0762 kg

b)

concentration of molality = 0,41mol / 1,0762 kg = 0,381 

ANSWER: concentration of molality = 0,381