Question

In how many grams of water should 25.31 g of potassium nitrate (kno3) be dissolved to prepare a 0.1982 m solution?

Answer 1

Answer: The mass of water that is needed to dissolve the given amount of potassium nitrate is 1263.1g

Explanation:

To calculate the mass of solvent for given molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Molality of the solution = 0.1982 m

$m_{solute}$ = Given mass of solute $(KNO_3)$ = 25.31 g

$M_{solute}$ = Molar mass of solute $(KNO_3)$ = 101.1 g/mol

$W_{solvent}$= Mass of solvent (water) = ?

Putting values in above equation, we get:

$0.1982=\frac{25.31\times 1000}{101.1\times \text{Mass of water}}\\\\\text{Mass of water}=\frac{25.31\times 1000}{101.1\times 0.1982}=1263.1g$

Hence, the mass of water that is needed to dissolve the given amount of potassium nitrate is 1263.1g

Answer 2

Solution:

Molality:-It is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent

Therefore,  

Molality = moles solute / kg solvent  

Therefore kg solvent = moles solute / molality  

moles solute = mass / molar mass  

= 25.31 g / 101.1 g/mole  

= 0.2503 mole

kg solvent = 0.2503 mole / 0.1982 m  

= 1.263 kg  

= 1263 g

This is the required answer.