How many grams of silver nitrate are needed to prepare 0.125 M solution in 250.0 mL of water?* (1 Point)
AgNO₃
Ag
= 107.868 8
mol
N = 14.0078
mol
8
0 = 15.999
mol
0.313 g
169.872 g
0.0849 g
5.31 g
O 31.3 g

5.3 grams AgNO3

Explanation:

To solve your problem, take a look at the definition of molality (m):

moles of solute/kg of solvent

You are given the grams of solvent (water) to find the grams of silver nitrate.

First, convert 250 grams to kg:

250 g * 1 kg / 1000g = 0.250 kg

Now that you have the kg of solvent, you can solve for the moles of solute:

kg of solvent * molality = moles of solute

0.250 kg * 0.125 m = 0.03125 mol AgNO3

You can convert the number of moles to grams using the molar mass of AgNO3, which is 169.872 g/mol:

moles of solute * molar mass of solute = grams of solute

0.03125 mol AgNO3 * 169.872 g/mol = 5.3085 g

Because your question provides two significant digits, you must round this number to get the final

5.3 grams AgNO3

I hope this helps!

the answer is b. dilution

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3