A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The
ID: 903449 • Letter: A
Question
A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting solution required 45.80 mL of 0.08765 M AgNO3(aq) to precipitate the Cl–(aq) and Br–(aq) as AgCl(s) and AgBr(s). Calculate the mass percentage of NaCl(s) in the mixture.
A 0.3146-g sample of a mixture of NaCI(s) and KBr(s) was dissolved in water. The resulting solution required 45.80 mL of 0.08765 M AgNOs(aq) to precipitate the Cr(aq) and Br (aq) as AgCI(s) and AgBr(s) Calculate the mass percentage of NaCI(s) in the mixture. Number % NaClExplanation / Answer
total mass of sample = 0.3146 grams
total mass of sample = mass of NaCl + KBr
No of moles of AgNo3 reacted = 45.8/1000*0.08765 = 0.0040144 mole
No of moles of AgNo3 reacted = total no of moles of NaCl + KBr = 0.0040144 mole
no of moles = weight/Mwt
total no of moles of NaCl + KBr = ((0.3146-x)/58.5)+(x/119.002) = 0.0040144
x = 0.157 grams
mass of NaCl = 0.3146-0.157 = 0.157 grams
mass percentage of NaCl = 0.157/0.3146*100 = 49.9%
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