What is the silver concentration in a solution prepared by mixing 250
ml of 0.310 m silver nitrate with 100 ml of 0.160 m of sodium chromate?

0.221M

Explanation:

From the question ,

The Molarity of AgNO₂ = 0.310 M

Hence , the concentration of Ag⁺ = 0.301 M

The volume of  AgNO₂  = 250 mL

and,

The Molarity of Sodium chromate = 0.160 M

The volume of Sodium chromate =  100 mL.

As the solution is mixed the final volume becomes ,

250mL +100mL = 350mL

Now, using the formula , to find the final molarity of the mixture ,

M₁V₁ ( initial ) =  M₂V₂ ( final )

substituting the values , in the above equation ,  

0.310M  *  250ml  =  M₂ * 350ml

M₂ = 0.221M

Hence , the concentration of the silver in the final solution = 0.221M

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explanation:

explanation:

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