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

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 :

$Molality=\frac{w_b}{M_b\times w_a}\times 1000$

where,

Molality = ?

$w_a$ = mass of solvent (water) = $2.00\times 10^{3}g$

$w_b$ = 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.

$Molality=\frac{373.5g}{331.2g/mole\times 2.00\times 10^{3}g}\times 1000$

$Molality=0.564mole/Kg$

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