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

2 answers:

8 0

Result: The amount of water required to dissolve the specified quantity of potassium nitrate is 1263.1g.

Clarification:

To determine the mass of solvent required for a specified molality, we utilize the formula:

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

where,

Molality of the solution = 0.1982 m

$m_{solute}$ = 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) =?

Substituting values into the equation yields:

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

Consequently, the necessary mass of water to dissolve the provided amount of potassium nitrate is 1263.1g.

6 0

Solution:

Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.

Thus,

Molality = moles of solute / kg of solvent.

Therefore, kg of solvent = moles of solute / molality.

moles of solute = mass / molar mass

= 25.31 g / 101.1 g/mole

= 0.2503 mole.

kg of solvent = 0.2503 mole / 0.1982 m

= 1.263 kg

= 1263 g.

This is the final answer.