# How many grams of AgNO3 are needed to make 250 ml of a solution that is 0 145 m?

To determine the grams of AgNO3 (silver nitrate) required to make 250 ml of a 0.145 m AgNO3 solution, we need to use the equation:

grams of solute = (molarity of solution) x (liters of solution) x (molar mass of solute)

Where:
– Molarity (m) is the concentration of the solution expressed in moles of solute per liter of solution
– Liters (L) is the volume of solution to be prepared
– Molar mass is the grams per mole of the solute

So for our problem:
– Molarity (m) = 0.145 m
– Liters (L) = 0.250 L
– Molar mass of AgNO3 = 169.87 g/mol

Plugging this into our equation:

Grams of AgNO3 = (0.145 m) x (0.250 L) x (169.87 g/mol)

= 6.24 g

So to make 250 ml of a 0.145 m AgNO3 solution, we would need 6.24 g of AgNO3.

## Step-by-Step Calculation

Here is a step-by-step way to calculate the grams of AgNO3 needed:

1. Write out the known values:
– Concentration (molarity) of AgNO3 solution: 0.145 m
– Volume of solution to prepare: 250 ml = 0.250 L
– Molar mass of AgNO3: 169.87 g/mol

2. Convert ml to L:
– 250 ml x (1 L / 1000 ml) = 0.250 L

3. Rearrange the equation:
grams of solute = (molarity) x (liters of solution) x (molar mass)

grams of AgNO3 = ?
Molarity of AgNO3 solution = 0.145 m
Liters of solution = 0.250 L
Molar mass of AgNO3 = 169.87 g/mol

4. Plug the values into the equation:

Grams of AgNO3 = (0.145 m) x (0.250 L) x (169.87 g/mol)
= 6.24 g

Therefore, to make 250 ml of a 0.145 M AgNO3 solution, 6.24 g of AgNO3 is needed.

## Explanation

To explain further:

– Molarity (M) is a concentration unit that measures the number of moles of solute per liter of solution.
– The molarity given in the problem, 0.145 m, tells us that in every 1 liter of solution, there are 0.145 moles of AgNO3.
– The volume given, 250 ml, is converted to liters to match the units of molarity.
– Using the molar mass of AgNO3 (169.87 g/mol), we can convert the moles of solute to grams.
– The molar mass is the grams per mole – it is the mass associated with 1 mole of the compound.
– So if we multiply the moles of AgNO3 by the molar mass, it gives us the mass in grams of AgNO3.
– This allows us to calculate the number of grams needed for the desired molarity and volume.

## Why Use Molarity?

Molarity (M) is commonly used to designate the concentration of a solution because:

– It directly relates moles of solute to volume of solution.
– The units work out conveniently to give grams when multiplied by molar mass.
– Molarity specifications allow solution concentrations to be standardized.
– Using molarity allows chemists to easily calculate amounts required for chemical reactions involving the solution.
– Molarity and moles are linked through Avogadro’s number, giving an exact known quantity of solute particles in solution.

Some examples of when molarity is useful:

– Following recipe-like procedures that require specific concentrations.
– Preparing reagents for experiments and assays.
– Performing titrations that rely on standardized concentrations.
– Expressing solution concentrations for chemical literature and communication.
– Performing stoichiometric calculations using molar ratios.

So being given the molarity makes it easy to calculate the mass of AgNO3 needed to prepare the desired volume. The molarity provides an exact concentration specification for the solution.

## Grams to Moles Conversion

The grams of AgNO3 can also be converted into moles for more insight:

– Grams of AgNO3 calculated: 6.24 g
– Molar mass of AgNO3: 169.87 g/mol
– Use dimensional analysis to convert:

6.24 g AgNO3 x (1 mol AgNO3 / 169.87 g AgNO3) = 0.0368 mol AgNO3

– So 6.24 g AgNO3 equals 0.0368 moles of AgNO3

This further demonstrates the relationship between mass in grams and amount in moles. The molar mass acts as the conversion factor.

Being able to interconvert between mass and moles is an extremely useful skill in chemistry calculations. The number of moles indicates the number of particles, which relates to the concept of Avogadro’s number (6.02 x 1023 particles/mol).

## Relationship Between Molarity and Moles

We can also explicitly show the relationship between molarity and moles:

– Molarity of AgNO3 solution = 0.145 m
– Volume of solution = 0.250 L
– Molarity x Volume = Moles

0.145 m x 0.250 L = 0.0363 moles AgNO3

– This matches our moles calculation from the gram amount.

– This makes sense because molarity gives moles/L, so:
(Molarity) x (Liters) = Moles

Being able to go between molarity, volume, grams and moles is important for solving a variety of chemistry stoichiometry and concentration problems.

## Summary

– To make 250 ml of 0.145 M AgNO3 solution requires 6.24 g of AgNO3
– Used the equation:
Grams = (Molarity)(Volume)(Molar mass)
– Molarity indicates moles/L, which can be converted to grams using the molar mass
– Knowing molarity allows calculation of the grams needed to prepare a specific volume and concentration
– Molarity, volume, grams and moles are connected quantities for expressing solution concentrations

## Practice Problems

Here are some additional practice problems to work on using these concepts:

1) How many grams of NaCl are needed to make 500 mL of a 0.15 M NaCl solution?

2) What volume of 0.25 M HCl solution can be made with 45 g of HCl?

3) If you have 125 mL of 0.5 M NaOH solution, how many moles of NaOH does it contain?

4) What is the molarity of a solution made by dissolving 5.85 g of MgCl2 in 125 mL of water?

Work through these practice problems to improve your skills in interconverting molarity, volume, grams and moles. Check your work using dimensional analysis and unit cancellation.

## Conclusion

– Molarity provides a convenient way to designate solution concentrations.
– Knowing the molarity allows calculation of the solute mass needed for a given solution volume.
– Conversion factors between molarity, volume, moles and mass enable solving for any desired quantity.
– Practicing these calculations is key for chemistry mastery.
– Understanding solution concentrations and being able to calculate between related quantities is an essential skill in chemistry.