Question

A solution is made by dissolving 373.5 g of Pb(NO3)2 (molar mass: 331.2 g/mol) in 2.00 × 103 g of water. What is the molality of the solution?

Answer 1

Molality = moles Pb(NO3)2/ Kg H2O

373.5 g Pb(NO3)2 X (1 mole/ 331.2 grams)= 1.13 moles Pb(NO3)2

2.00 x 10^3 g (1 Kg/ 1000g)= 2 Kg

Molality= 1.13 moles/ 2 Kg= 0.565 molal.

Answer 2

Answer : The molality of the solution is, 0.564 mole/Kg

Step-by-step explanation:

Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

In this question, the solute is
`Pb(NO_3)_2` and solvent is water.

Formula used :

`Molality=(w_b)/(M_b* w_a)* 1000`

where,

Molality = ?

`w_a` = mass of solvent (water) =
`2.00* 10^(3)g`

`w_b` = mass of solute
`Pb(NO_3)_2` = 373.5 g

`M_b` = molar mass of solute
`Pb(NO_3)_2` = 331.2 g/mole

Now put all the given values in the above formula, we get the molality of the solution.

`Molality=(373.5g)/(331.2g/mole* 2.00* 10^(3)g)* 1000`

`Molality=0.564mole/Kg`

Therefore, the molality of the solution is, 0.564 mole/Kg