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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?

Respuesta :

Ankit

Answer:

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

mahima6824soni

1. the grams of [tex]\rm AgNO_3[/tex] 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 [tex]\rm AgNO_3[/tex]

[tex]\rm Molarity = \dfrac{n}{V}\\\\rm 6 = \dfrac{n}{1}\\\\n = 6 \times 1 = 6[/tex]

Mass of One mole of [tex]\rm AgNO_3[/tex] 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

[tex]\rm density = \dfrac{mass }{volume} \\\\\rm density = \dfrac{1019.22 }{0.1} = 10,192.2[/tex]

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

Learn more about molarity, here

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