0.550g sample containing only nacl and kbr.it dissolved in water and titrated with a 0.0322M solution of agno3. complete titration of both halides was found to require 30.45ml of agno3 solution. calculate the weight percent of cl in the solid sample.

Please note that nacl doesn't mean anything. kbr is nothing either. cl is nothing. Starting a sentence with a small letter makes the sentence difficult to read and interpret. Henceforth, please write your questions in good style. I'm helping because I managed to understand the question after two or three attempts but MOST times I let the question go unanswered.

Set the problem up as two simultaneous equations.
Let X = mass NaCl
let Y = mass KCl
=================
X + Y = 0.550 (equation 1).
moles X + moles Y = moles AgNO3.
(X/molar mass NaCl) + (Y/molar mass KCl) = moles AgNO3. (equation 2).
moles AgNO3 = M x L.
The problem asks for weight percent Cl. Therefore, convert mass NaCl to mass Cl, then percent Cl = (mass Cl/0.550)*100 = ??
Post your work if you get stuck.

To find the weight percent of Cl in the solid sample, we need to calculate the moles of Cl and the total mass of the sample.

1. Calculate the moles of AgNO3 used in the titration:
Molarity (M) = moles/volume (L)
0.0322 M = moles/0.03045 L
Moles of AgNO3 = 0.0322 M * 0.03045 L = 0.0009824 moles

2. Calculate the moles of Cl in the sample using the stoichiometry of the reaction:
From the balanced equation: AgNO3 + NaCl -> AgCl + NaNO3
1 mole of AgNO3 reacts with 1 mole of NaCl, which means 0.0009824 moles of AgNO3 react with 0.0009824 moles of NaCl

3. Calculate the mass of Cl in the sample:
Molar Mass of NaCl = atomic mass of Na + atomic mass of Cl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Mass of Cl = Moles of Cl * Molar Mass of Cl = 0.0009824 moles * 35.45 g/mol = 0.0347843 g

4. Calculate the weight percent of Cl:
Weight percent = (Mass of Cl / Total mass of sample) * 100
Total mass of sample = 0.550 g
Weight percent of Cl = (0.0347843 g / 0.550 g) * 100 = 6.33%

Therefore, the weight percent of Cl in the solid sample is approximately 6.33%.

To calculate the weight percent of Cl in the solid sample, we first need to determine the number of moles of Cl in the sample.

Given:
Mass of the sample = 0.550 g
Volume of AgNO3 solution used for titration = 30.45 mL
Molarity of AgNO3 solution = 0.0322 M

Step 1: Calculate the number of moles of AgNO3 used in the titration.
Moles of AgNO3 = Volume (L) x Molarity
Convert the volume from mL to L:
Volume of AgNO3 used in titration = 30.45 mL * (1 L/1000 mL) = 0.03045 L
Moles of AgNO3 = 0.03045 L x 0.0322 M = 0.000982 moles

Step 2: Calculate the number of moles of Cl in the sample.
Since AgNO3 reacts with Cl to form AgCl, and the reaction is 1:1, the moles of Cl will be equal to the moles of AgNO3 used in the titration.
Moles of Cl = 0.000982 moles

Step 3: Calculate the weight of Cl in the sample.
Since the molar mass of NaCl is 58.44 g/mol, we can calculate the weight of Cl using the mole-to-mass ratio.
Weight of Cl = Moles of Cl x Molar mass of Cl
Weight of Cl = 0.000982 moles x (35.45 g/mol) = 0.0348 g

Step 4: Calculate the weight percent of Cl in the sample.
Weight percent of Cl = (Weight of Cl / Total sample weight) x 100
Weight percent of Cl = (0.0348 g / 0.550 g) x 100 = 6.33 %

Therefore, the weight percent of Cl in the solid sample is approximately 6.33%.