A 122mL sample of a 1.2 M sucrose solution is diluted to 500.0 mL. what is the molarity of the diluted solution

The molarity of the diluted solution is 0.29 M

From the question,

We are to determine the molarity of the diluted solution.

Using the dilution law formula

M₁V₁ = M₂V₂

Where

M₁ is the initial molarity

V₁ is the initial volume

M₂ is the final molarity

and V₂ is the final volume

From the given information,

M₁ = 1.2 M

V₁ = 122 mL

V₂ = 500.0 mL

Putting the parameters into the formula, we get

∴

∴ M₂ = 0.2928 M

M₂ ≅ 0.29 M

Hence, the molarity of the diluted solution is 0.29 M

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0.30M

Explanation:

Let solution 1 be the concentrated solution and solution 2 be the diluted solution

(Molarity x Volume)₁ = (Molarity x Volume)₂

(122ml)₁·(1.2M)₁ = (M)₂·(500ml)₂ => M₂ = 122ml ·1.2M / 500ml = 0.2928M ≅ 0.30M

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

i believe the answer is a.