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"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute this 25.0 mL sample with water until the new volume equals 600.0 mL, what is the concentration of nitrate ions in the 600.0 mL solution

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emaikwuvictor

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

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