1, When 4.0 mole of LiI is dissolved 1000.0 ml of water, what resulting molality

Question

1, When 4.0 mole of LiI is dissolved 1000.0 ml of water, what resulting molality (m) is used to calculate the change in boiling point of the water.
2, What mass ( in g) of NH3 must be dissolved in 1000.0 g of Methanol to make a 0.250m solution?

1, When 4.0 mole of LiI is dissolved 1000.0 ml of water, what resulting molality (m) is used to calculate the change in boiling point of the water.
2, What mass ( in g) of NH3 must be dissolved in 1000.0 g of Methanol to make a 0.250m solution?

1, When 4.0 mole of LiI is dissolved 1000.0 ml of water, what resulting molality (m) is used to calculate the change in boiling point of the water.
2, What mass ( in g) of NH3 must be dissolved in 1000.0 g of Methanol to make a 0.250m solution?

Explanation / Answer

1) molality = moles of solute / mass of solvent in kg

   Given that moles of LiI = 4 .0 mol

              Since density of water is 1 g/mL,

           mass of water = 1000 g = 1 kg

Hence,

   molality = moles of LiI / mass of water in kg = 0.4 mol/1 kg = 0.4 m

Therefore, resulting molality = 0.4 m

2)   molality = moles of solute / mass of solvent in kg

   molality = ( mass / molar mass) x (1/ mass of solvent in kg)

          Given that molality = 0.25 m

                     mass of solvent methanol = 1000 g = 1kg

                molar mass of NH3 = 17 g/mol

   Then,

molality = ( mass of NH3/ molar mass of NH3) x (1/ mass of solvent in kg)

0.25 m = ( mass of NH3/ 17 g/mol) x ( 1/1 kg)

mass of NH3 = 4.25 g

   Therefore,

    mass of NH3 to be dissolved = 4.25 g

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.