Answer to Question #174529 in General Chemistry for lesego

Question

a 5.000g sample a pesticide was decomposed with metallic sodium in alcohol, and the liberated chloride ion was precipitated as agcl. express the result of tis analysis in terms of percentage ddt based upon the recovery of 0.1606 g of agcl.

Answer 1

Molecular weight of AgCl = 107.87 + 35.45 = 143.42 g/mol.

0.1606 g AgCl = 0.1606 /143.42 mol = 0.00112 mol.

So, weight of Cl present = 0.00112 × 35.45 g

= 0.0397 g.

The ddt percentage = 0.0397/5.000 × 100%

= 0.79 %.