Question

What mass of $AgN{O}_3$ would be required to prepare a 0.250 molal solution in 125 g of water?

Answer 1

Answer
$5.31gAgN{O}_3$
Explanation :
Your strategy here will be to
Use the molarity and mass of water to calculate how many moles of solute you have in your solution
Use the molar mass of silver nitrate to convert the moles to grams
As you know , molality is defined as moles of solute, which in your case is silver nitrate, per kilogram of solvent . This means that a solution's molality essentially tells you the number of moles of solute present in 1 kg of solvent.
In your case, a 0.250 molal solution would contain 0.250 moles of solute for every 1 kg of solvent
The problem tells you that you have 125 g of solvent available. Use the solution's molality as a conversion factor to calculate how many moles of silver nitrate it must contain -- do not forget to convert the mass of solvent from grams to kilograms
Now, you know that silver nitrate has a molar mass of $169.87gmo{l}^{-1}$ , which means that one mole of silver nitrate has a mass of 169.87 g.
You can thus say that 0.03125 moles of silver nitrate will have a mass of
$molesAgN{O}_3\frac{169.87g}{1moleAgN{O}_3}=5.31g$
The answer is rounded to three sig fig