Question

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1. How many grams of AgNO3 are necessary to make 1.0 L of a 6.0 M stock solution?

2. How would you make 1.0 L of a 0.1 M solution of AgNO3 from your 6.0 M stock solution?

Answer 1

1. the grams of
`\rm AgNO_3` is 1019.22.

2. 10,192.2

What is molarity?

Molarity is the measure of the concentration of any solute in per unit volume of the solution.

1. Volume is 1.0 l.

Molarity of solution -6 m

To find the mass of
`\rm AgNO_3`

`\rm Molarity = (n)/(V)\\\\rm 6 = (n)/(1)\\\\n = 6 * 1 = 6`

Mass of One mole of
`\rm AgNO_3` is 169.87 g

Therefore, the mass of 6 moles will be

169.87 × 6 = 1019.22

2. Molarity of solution 6.0

Volume of solution is 0.1m

The mass of 6 moles will be

169.87 × 6 = 1019.22

`\rm density = (mass )/(volume) \\\\\rm density = (1019.22 )/(0.1) = 10,192.2`

Thus, the options are 1. 1019.22 2. 10,192.2

Learn more about molarity, here

Answer 2

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution