A solution contains 0.253 g of calcium cyanide in 2.00 L. A . What is the concen

Question

A solution contains 0.253 g of calcium cyanide in 2.00 L. A . What is the concentration of this solution in parts per million ?

B. What is the concentration of cynaide ion in this solution ?


C. How many mL of this solution must be diluted to prepare 250.mL of a solution with cyanide ion concentration of 15 ppm? A solution contains 0.253 g of calcium cyanide in 2.00 L. A . What is the concentration of this solution in parts per million ?

B. What is the concentration of cynaide ion in this solution ?


C. How many mL of this solution must be diluted to prepare 250.mL of a solution with cyanide ion concentration of 15 ppm? A . What is the concentration of this solution in parts per million ?

B. What is the concentration of cynaide ion in this solution ?


C. How many mL of this solution must be diluted to prepare 250.mL of a solution with cyanide ion concentration of 15 ppm?

Explanation / Answer

Ca(CN)2 will give 2 CN- ions in solution.

A. Now 1mg/L = 1 ppm

0.253 g = 0.253 x 1000 = 253 mg

So concentration in mg/L = 253 mg/2L = 126.5 mg/L = 126.5 ppm

B) since concentration of cyanide ion will double, therefore it's concentration will be 126.5 x 2 = 253 ppm

C) c1 x v1 = c2 x v2

Therefore 253 ppm x v1 mL = 15 ppm x 250 mL

So v1 = (15 x 250)/253 = 14.8 mL

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