Question

644 g of na2so4.10h2o is dissolved in 1640 g of water then molality of solution is

Answer 1

Molality = 644/322×1000/1640

= 1.22 molal.

Answer 2

Answer:

The molality of the given solution measured is $1.219 moles/Kg$.

Explanation:

Given,

The mass of the solute, i.e., sodium sulfate decahydrate, $Na_{2}SO_{4}.10H_{2}O$ = $644g$

The mass of the solvent, i.e., water = $1640 g$ = $1.640 Kg$

The solution's molality, m =?

As we know,

• Molality is defined as the moles of a solute per kilogram of the solvent.
• $Molality = \frac{Total moles of solute}{Mass of solvent (Kg)}$    -------equation (1)

Now, we have to find out the moles of the solute by the equation given below:

• Number of moles of solute = $\frac{Mass of solute}{Molar mass of solute}$

Here,

• The molar mass of $Na_{2}SO_{4}.10H_{2}O$ = $322g/mol$

Therefore,

• Number of moles of solute = $\frac{644}{322}$ = $2 moles$

After putting the value of the number of moles of solute in equation (1), we get:

• $Molality = \frac{Total moles of solute}{Mass of solvent (Kg)}$
• Molality = $\frac{2}{1.640}$ = $1.219 moles/Kg$

Hence, the solution's molality, m = $1.219 moles/Kg$.