Question

A 0.750 g sample containing only KCl and NaBr was titrated with 0.1043 M AgNO3 , requiring 42.23 mL to reach the endpoint. Another sample have the same mass was taken and the halogen(Br and Cl) was precipitated as AgCl and AgBr using the same solution of AgNO3, the precipitated mass was found to be 0.8042 gm. ( What is the % NaBr in the sample?

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2 points
53.35 %
46.65%
60 %
non of all

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the % NaBr in the sample, we first need to calculate the moles of AgNO3 used in the titration.

Moles of AgNO3 = 0.1043 mol/L * 0.04223 L = 0.00440119 mol

Since AgNO3 reacts in a 1:1 ratio with the Cl- and Br- ions in NaBr and KCl, we can deduce that the moles of Cl- and Br- ions in the sample are equal to the moles of AgNO3 used in the titration.

Moles of Cl- and Br- ions = 0.00440119 mol

Now, we need to calculate the moles of AgCl and AgBr formed in the second sample.

Moles of AgCl + AgBr = 0.8042 g / (143.32 g/mol + 187.77 g/mol) = 0.0044818 mol

Since AgCl and AgBr form in a 1:1 ratio with Cl- and Br- ions respectively, the moles of Cl- and Br- ions in the sample can be calculated by dividing the total moles of AgCl and AgBr by 2.

Moles of Cl- and Br- ions = 0.0044818 mol / 2 = 0.0022409 mol

Now, we can calculate the moles of NaBr in the sample by subtracting the moles of Cl- ions from the total moles of Cl- and Br- ions.

Moles of NaBr = Moles of Cl- and Br- ions - Moles of Cl- ions
Moles of NaBr = 0.00440119 mol - 0.0022409 mol = 0.00216029 mol

Next, we can calculate the mass of NaBr in the sample.

Mass of NaBr = Moles of NaBr * (22.99 g/mol + 79.90 g/mol)
Mass of NaBr = 0.00216029 mol * 102.89 g/mol = 0.22232 g

Finally, we can calculate the % NaBr in the sample.

% NaBr = (Mass of NaBr / Total sample mass) * 100
% NaBr = (0.22232 g / 0.750 g) * 100 = 29.64%

Therefore, the % NaBr in the sample is approximately 29.64%.