Question

Suppose 18.3 g of sodium iodide is dissolved in 300 mL of a 0.30 Maqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Be sure your answer has the correct number of significant digits. X 52

Answer 1

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As we have 18.3 gm of No I 2 Its M. W = 149.89 gm mole So moles of NaI = 18.3gm 144.89 gm/male = 0.122 mole Now Ag NO3 + Na I - Ag ICs) + Na NO3 Se solid Ag I formed will settle down a no I will be in solo Now Malarity = 0. 3M 2 Valume = 300 ml or o. 3 Ļ moles of AgNO3 = olarity X Ven 2 TAS IL=1000mL) = 0.3 x 0.3=0.09 #so Settle out of 0.122 Nol 0.09 will get down. So I o in sal» - [0.122-0009] = 100032 Malarity = no of moles - 0.032 vin Litre 0.3L Malarity = 00107