Question

"what is the molality of a solution containing 1.567 grams of molten iron (fe) solvent and 0.0021 grams of molten lead (pb) solute?"

Answer 1

the molality of a solution containing is
$6 \times 10 { - 6}^{?}$

Answer 2

Answer : The molality of solution is, $6.467\times 10^{-3}mole/Kg$

Solution : Given,

Mass of solute = 0.0021 g

Mass of solvent = 1.567 g

Molar mass of Pb = 207.2 g/mole

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

Formula used :

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

where,

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

Now put all the given values in the formula of molarity, we get

$Molality=\frac{0.0021g\times 1000}{207.2g/mole\times 1.567Kg}=6.467\times 10^{-3}mole/Kg$

Therefore, the molality of solution is, $6.467\times 10^{-3}mole/Kg$