What is the molarity of an aqueous solution that contains 12.2 grams of srcl2 dissolved in 2,500 milliliters of solution?

Answers

Molar mass SrCl₂ = 158.53 g/mol

number of moles:

mass SrCl₂ / molar mass

12.2 / 158.53 => 0.0769 moles

2,500 mL in liters: 2,500 / 1000 => 2.5 L

Therefore:

M = moles / volume

M = 0.0769 / 2.5

M = 0.03076 mol/L⁻¹

0.031 M is the molarity of an aqueous solution that contains 12.2 grams of SrCl2 dissolved in 2,500 milliliters of solution.

0.031 M is the molarity of an aqueous solution that contains 12.2 grams of SrCl2 dissolved in 2,500 milliliters of solution.

Molarity is

mol

Litre 

First you need to find how many moles 12.2 grams is 

To do this you need to find molecular weight of SrCl2
(87.62)+(35.45*2) = 158.52 g/mol

Now convert 12.2 g into moles
12.2 g * 1 mol / 158.52 g (12.2/158.52) = 0.0769618976 moles

Next convert 2500 mL into Litres
2500mL * 1L / 1000mL (2500/1000) = 2.5 Litres

Finally do moles / Litres

0.0769618976 / 2.5 = 0.0308 M
Molarity is

mol

Litre 

First you need to find how many moles 12.2 grams is 

To do this you need to find molecular weight of SrCl2
(87.62)+(35.45*2) = 158.52 g/mol

Now convert 12.2 g into moles
12.2 g * 1 mol / 158.52 g (12.2/158.52) = 0.0769618976 moles

Next convert 2500 mL into Litres
2500mL * 1L / 1000mL (2500/1000) = 2.5 Litres

Finally do moles / Litres

0.0769618976 / 2.5 = 0.0308 M


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