Question

58.0 g of K2SO4 was dissolved in 500 g of water. What is the molality of this solution?

Answer 1

Answer: The molality of solution is 0.66 mole/kg

Step-by-step explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

`Molarity=(n* 1000)/(W_s)`

where,

n = moles of solute

`W_s` = weight of solvent in g

moles of
`K_2SO_4` =
`\frac{\text {given mass}}{\text {Molar Mass}}=(58.0g)/(174g/mol)=0.33mol`

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

`Molality=(0.33* 1000)/(500g)=0.66mole/kg`

Therefore, the molality of solution is 0.66 mole/kg