Question

what mass of the fallowing solutions contains 0.100 moles of solute?

A.0.300 m NH4NO3

B. 1.85 m ethylene glycol (C2H6O2)

C. 5.13 m CaCl2

Answer 1

A) we know that

molality = moles of solute / mass of solvent (kg)

so

0.3 = 0.1 / mass of solvent (kg)

mass of solvent (kg) = 0.333 kg = 333 g

now

mass of solute = moles of soltue x molar mass

so

mass of solute = 0.1 x 80 = 8 g

mass of solute = 8 g

so

total mass of solutieon = 8 + 333

total mass of solution = 341 g

B)

we know that

molality = moles of solute / mass of solvent (kg)

so

1.85 = 0.1 / mass of solvent (kg)

mass of solvent (kg) = 0.054 kg = 54 g

now

mass of solute = moles of soltue x molar mass

so

mass of solute = 0.1 x 62 = 6.2 g

mass of solute = 6.2 g

so

total mass of solutieon = 6.2 + 54

total mass of solution = 60.2 g

C)

we know that

molality = moles of solute / mass of solvent (kg)

so

5.13 = 0.1 / mass of solvent (kg)

mass of solvent (kg) = 0.0195 kg = 19.5 g

now

mass of solute = moles of soltue x molar mass

so

mass of solute = 0.1 x 11 = 11.1 g

mass of solute = 11.1 g

so

total mass of solutieon = 11.1 + 19.5

total mass of solution = 30.6 g