Class 12 Exam > Class 12 Questions > A solution containing 3.5 g of solute X is 50...
Start Learning for Free
A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)
- a)molality of the solution is 0.46 mol kg-1
- b)molarity of the solution is 0.44 mol L-1
- c)molar mass of solute X is 152 g mol-1
- d)mole fraction of solute X is 0.0082
Correct answer is option 'A,B,C,D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A solution containing 3.5 g of solute X is 50.0 g of water has a volum...
(b) Total mass of solution = 50 .0 + 3.5 Volume of solution = 52 .5 m L
(c) 70 g X = 0 .4 6 mol (present in 1000 g H20 )
Most Upvoted Answer
A solution containing 3.5 g of solute X is 50.0 g of water has a volum...
Given:
- Mass of solute X (mX) = 3.5 g
- Mass of water (mH2O) = 50.0 g
- Volume of solution (V) = 52.5 mL = 52.5 cm³
- Freezing point depression (ΔTf) = -0.86 °C
- Freezing point constant (Kf) of water = 1.86 mol⁻¹ kg
To Find:
a) Molality of the solution
b) Molarity of the solution
c) Molar mass of solute X
d) Mole fraction of solute X
Solution:
a) Molality (m):
Molality is defined as the number of moles of solute per kilogram of solvent.
Molality (m) = (moles of solute X) / (mass of solvent in kg)
Given:
Mass of solute X (mX) = 3.5 g
Mass of water (mH2O) = 50.0 g
First, we need to convert the mass of water to kg:
Mass of water (mH2O) = 50.0 g = 0.050 kg
Now, we can calculate the molality:
Molality (m) = (moles of solute X) / (mass of solvent in kg)
To find the moles of solute X, we need to know its molar mass.
b) Molarity (M):
Molarity is defined as the number of moles of solute per liter of solution.
Molarity (M) = (moles of solute X) / (volume of solution in L)
Given:
Volume of solution (V) = 52.5 mL = 52.5 cm³ = 0.0525 L
Now, we can calculate the molarity:
Molarity (M) = (moles of solute X) / (volume of solution in L)
To find the moles of solute X, we need to know its molar mass.
c) Molar Mass of Solute X:
Molar mass is defined as the mass of one mole of a substance.
Molar mass of solute X (Molar mass) = (moles of solute X) / (mass of solute in g)
Given:
Mass of solute X (mX) = 3.5 g
Now, we can calculate the molar mass:
Molar mass of solute X (Molar mass) = (moles of solute X) / (mass of solute in g)
d) Mole Fraction of Solute X:
Mole fraction is defined as the ratio of the number of moles of one component to the total number of moles in the solution.
Mole fraction of solute X = (moles of solute X) / (moles of solute X + moles of solvent H2O)
Given:
Mass of solute X (mX) = 3.5 g
Mass of water (mH2O) = 50.0 g
First, we need to convert the masses to moles:
Moles of solute X = (mass of
- Mass of solute X (mX) = 3.5 g
- Mass of water (mH2O) = 50.0 g
- Volume of solution (V) = 52.5 mL = 52.5 cm³
- Freezing point depression (ΔTf) = -0.86 °C
- Freezing point constant (Kf) of water = 1.86 mol⁻¹ kg
To Find:
a) Molality of the solution
b) Molarity of the solution
c) Molar mass of solute X
d) Mole fraction of solute X
Solution:
a) Molality (m):
Molality is defined as the number of moles of solute per kilogram of solvent.
Molality (m) = (moles of solute X) / (mass of solvent in kg)
Given:
Mass of solute X (mX) = 3.5 g
Mass of water (mH2O) = 50.0 g
First, we need to convert the mass of water to kg:
Mass of water (mH2O) = 50.0 g = 0.050 kg
Now, we can calculate the molality:
Molality (m) = (moles of solute X) / (mass of solvent in kg)
To find the moles of solute X, we need to know its molar mass.
b) Molarity (M):
Molarity is defined as the number of moles of solute per liter of solution.
Molarity (M) = (moles of solute X) / (volume of solution in L)
Given:
Volume of solution (V) = 52.5 mL = 52.5 cm³ = 0.0525 L
Now, we can calculate the molarity:
Molarity (M) = (moles of solute X) / (volume of solution in L)
To find the moles of solute X, we need to know its molar mass.
c) Molar Mass of Solute X:
Molar mass is defined as the mass of one mole of a substance.
Molar mass of solute X (Molar mass) = (moles of solute X) / (mass of solute in g)
Given:
Mass of solute X (mX) = 3.5 g
Now, we can calculate the molar mass:
Molar mass of solute X (Molar mass) = (moles of solute X) / (mass of solute in g)
d) Mole Fraction of Solute X:
Mole fraction is defined as the ratio of the number of moles of one component to the total number of moles in the solution.
Mole fraction of solute X = (moles of solute X) / (moles of solute X + moles of solvent H2O)
Given:
Mass of solute X (mX) = 3.5 g
Mass of water (mH2O) = 50.0 g
First, we need to convert the masses to moles:
Moles of solute X = (mass of
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer?
Question Description
A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer?.
A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer?.
Solutions for A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer?, a detailed solution for A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? has been provided alongside types of A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A solution containing 3.5 g of solute X is 50.0 g of water has a volume of 52.5 mL and a freezing point of - 0.86° C. Thus, (Kf of H20 = 1.86° mol-1 kg)a)molality of the solution is 0.46 mol kg-1b)molarity of the solution is 0.44 mol L-1c)molar mass of solute X is 152 g mol-1d)mole fraction of solute X is 0.0082Correct answer is option 'A,B,C,D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup