Question

suppose 3.50% of aqueous solution of NaCl is sea water what is the molality of seawater

Answer 1

as 3.5% of aqueous solution is common salt so 1000gms of the solution has 35 gms of solute so 965 gms of water has 35gms of NaCl

Answer 2

Answer : The molality of solution is 0.6 mol/kg.

Solution : Given,

3.50 g of NaCl is present in 100 g of water.

Molar mass of $NaCl$ = 58.5 g/mole

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$  

$Moles=\frac{3.50g}{58.5}=0.06moles$  

Molality : It is defined as the number of moles of solute present per kg of the solvent.

Formula used :

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute $NaCl$ = 0.06 moles

$W_s$ = Weight of solvent in kg = 100 g= 0.1 kg

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

$Molality=\frac{0.06moles}{0.1kg}=0.6mol/kg$

Therefore, the molality of solution is 0.6 mole/kg