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5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)?
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5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point...
Calculation of x in the given problem
Formula used:
ΔTf = Kf x molality
Explanation:
When Na2SO4 is dissolved in water, it dissociates into 2 Na+ ions and 1 SO4^2- ion. Therefore, the total number of particles formed in the solution is 3. Hence, the van't Hoff factor (i) is 3.
The molality (m) of the solution can be calculated using the formula:
m = (moles of solute) / (mass of solvent in kg)
We are given that 5g of Na2SO4 is dissolved in x g of H2O. The molar mass of Na2SO4 is 142g/mol. Therefore, the number of moles of Na2SO4 can be calculated as follows:
moles of Na2SO4 = (5g) / (142g/mol) = 0.0352 mol
Since Na2SO4 is 81.5% ionized, the actual number of moles of solute present in the solution will be:
moles of solute = (0.0352 mol) x (81.5/100) = 0.0287 mol
Now, we can calculate the molality of the solution as follows:
m = (0.0287 mol) / (x/1000 kg) = 28.7 (x/1000) mol/kg
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf x i x m
where Kf is the freezing point depression constant and i is the van't Hoff factor.
Substituting the values, we get:
3.82 = (1.86 CKg/mol) x 3 x 28.7 (x/1000) mol/kg
Solving for x, we get:
x = (3.82 x 1000) / (1.86 x 3 x 28.7) = 45.4 g
Therefore, 45.4 g of water is required to dissolve 5 g of Na2SO4 and produce a solution with a freezing point depression of 3.82°C.
Community Answer
5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point...
Got it... 45 grams (44.919)
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5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)?
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5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)?.
5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5g of Na2SO4 was dissolved in x g of H2O. The change in freezing point is 3.82 celcius. If Na2SO4 is 81.5 percent ionised then find x. (Kf is 1.86 CKg/mol, molar mass of S=32g Na=23g)?.
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