What is the mass of precipitate formed when 50mL of 16.9% solution of `AgNO_3` is mixed with 50 mL of 5.8% NaCI solution? 9Ag=107.8,N=14,O=16,Na=23CI=35.5)`

Options

A. 28g
B. 3.5 g
C. 7 g
D. 14 g

Answer

C

Explanation

(c ) 50 mL of 16.95 solution of `A_gNO_3(16.0/100x50)=8.45g of AgNO_3``n_(mol e)=(98.45 g)/((107.8+14+16xx3)g/(mol))=((8.45g)/9169.8g/(mol))=0.0497mol es` 50mL of 5.8% solution of NaCI contains `NaCI=((5.8)/(100)xx50)=2.9``NaCI=((5.8)/(100)xx50)=2.9``underset"1 mol e"(AgNO_3)"`+`underset"1 mol e"(NaCI)"``rarrunderset"1mol e"(AgCIdownarrow`+`underset"Na"oplus+underset"CI"oplus``therefore`0.049 mole 0.049 mole 0.049 mole of AgCI`n=w/Mimpliesw =(n_(AgCI)xx`molecular mass `=(0.049)xx91.07.8+35.5)=7.02g`.