The element Se, dispersed in 2 ml sample of detergent for dandruff control, was determined by suspending it in the warm ammoniacal solution that contained 45 ml of 0.02 M − AgN O 3 . 6A g + + 3Se(s)

Question

The element Se, dispersed in 2 ml sample of detergent for dandruff control, was determined by suspending it in the warm ammoniacal solution that contained 45 ml of 0.02 $M - AgN O_{3}$ . $6A g^{+}$ + 3Se(s) + $6N H_{3}$ + $3 H_{2} O$ $\to$ $2A g_{2} Se (s)$ + $A g_{2} Se O_{3} (s)$ + $6N H_{4}^{+}$ . The mixture was now treated with excess nitric acid which dissolved the $A g_{2} Se O_{3}$ but not $A g_{2} Se$ . The $A g^{+}$ from the $A g_{2} Se O_{3}$ and the excess of $AgN O_{3}$ consumed 30 ml of 0.01 N-KSCN in Volhard titration. How many milligrams of Se was contained per ml of sample? (Se = 80)

Answer:

18.00

Solution:

$AgN O_{3}$ taken = 0.9 mmol. For $n$ mmol Se, Ag in $A g_{2} Se$ = 4n/3. Remaining Ag = $30 \times 0.01$ = 0.3 mmol. So 0.9 = 4n/3 + 0.3, $n$ = 0.45 mmol = 36 mg in 2 ml. Per ml = 18 mg. Ans 0018.

TestHub · Accepted Answer

AgN O 3 ​ taken = 0.9 mmol. For n mmol Se, Ag in A g 2 ​ Se = 4n/3. Remaining Ag = 30 × 0.01 = 0.3 mmol. So 0.9 = 4n/3 + 0.3, n = 0.45 mmol = 36 mg in 2 ml. Per ml = 18 mg. Ans 0018.

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