Question
Question: How many sodium fluoride (NaF) are needed to prepare a 0.400 m NaF solution that contains 750 g of w...
How many sodium fluoride (NaF) are needed to prepare a 0.400 m NaF solution that contains 750 g of water?
Solution
We know that molality is also a way used to express concentration of the solution. Molality is defined as the moles of solute present in 1 kg of solvent. Here, we have to use the formula of molality, that is, Molality=MassofsolventinKgMolesofsolute.
Complete step by step answer:
Here, first we have to calculate the moles of NaF using the formula of molality. Then, we have to calculate the mass of NaF using the formula of moles.
Given, the molality of the solution is 0.400 m and the mass of the solvent is 750 g.
Now, we have to convert the mass of solvent from g to kg.
750g=1000750kg = 43kg
Now, we put the above values in the formula of molality.
Molality=MassofsolventinKgMolesofsolute
⇒0.400mkg−1=43kgMolesofNaF
⇒Moles of NaF=0.400×43=0.3
Now, we have to use the formula of moles to calculate the mass of NaF. The formula is,
Number of moles=MolarmassMass
We calculated the number of moles of NaF in the previous step. Now, we have to calculate the molar mass of NaF.
Molar mass of NaF=23+19=42gmol−1.
So,
Mass of NaF=Moles of NaF ×Molar mass of NaF
⇒ Mass of NaF=0.3×42=12.6g
Therefore, 12.6 g of NaF is needed to prepare a 0.400 m NaF solution that contains 750 g of water.
Note: It is to be noted that molarity and molality are not the same. Molarity is the moles of solute present in 1 litre of the solution and the molarity is the moles of solute present in 1kg of solvent. Symbolically molarity is represented by M and molality is represented by m.