Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by R.Sasi Kumar 2 years, 10 months ago
- 3 answers
Anjan Karthi 2 years, 10 months ago
Posted by Ananya Kasaudhan 2 years, 10 months ago
- 1 answers
Yashi Kushwaha 2 years, 8 months ago
Posted by K. Priyanka_Xi_C_16 2 years, 10 months ago
- 1 answers
Sushma Pandey 2 years, 10 months ago
Posted by Sazid Sheikh 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
-
- Due to greater electronegativity of sp2 hybridised carbon to which carboxyl carbon is attached and greater resonance stabilization of carboxylate ion with the benzene ring, benzoic acid is a stronger acid than acetic acid.
- Methanal is more reactive towards nucleophilic addition reaction than ethanal because carbonyl carbon of methanal is more electrophilic than that of ethanol. Also, due to +I effect of methyl group in ethanal, reactivity of ethanal decreases.
Posted by Priu S 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
The dual cell heat exchanger consists of two porous fluid zones, namely a primary zone and an auxiliary zone. The two zones are solved simultaneously and are coupled only through heat transfer.
Posted by Aniket . 2 years, 10 months ago
- 1 answers
Posted by Shalu Shrivastav 2 years, 10 months ago
- 0 answers
Posted by Shaurya Singh 2 years, 10 months ago
- 3 answers
Posted by Ahuja Yadav 2 years, 11 months ago
- 1 answers
Meemansha Srivastava 2 years, 11 months ago
Posted by Shubham Baba 2 years, 11 months ago
- 1 answers
Posted by Annu Kri 2 years, 11 months ago
- 2 answers
Anjan Karthi 2 years, 11 months ago
Shubham Baba 2 years, 11 months ago
Posted by Parth Vaishya 03 2 years, 11 months ago
- 1 answers
Posted by Jai Vignesh 2 years, 11 months ago
- 1 answers
Kashu Verma 2 years, 10 months ago
Posted by Madhwan Pratap 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
{tex}\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}{/tex}
Oxidation of one mole of water will require 2×96500=1.93×105C.
Posted by Farman Malik 2 years, 11 months ago
- 1 answers
Kanhaiya Saraswat 2 years, 11 months ago
Posted by Garima Sharma 2 years, 11 months ago
- 4 answers
Posted by Garima Sharma 2 years, 11 months ago
- 3 answers
Posted by Falak Vaishya 2 years, 11 months ago
- 1 answers
Anjan Karthi 2 years, 11 months ago
Posted by R.Sasi Kumar 2 years, 11 months ago
- 2 answers
Posted by Kashish Rathore 2 years, 11 months ago
- 1 answers
Posted by Avaneesh Rao 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.
Posted by Rajesh Patel Rajesh 2 years, 11 months ago
- 0 answers
Posted by Ashish Yadav 2 years, 11 months ago
- 3 answers
Naman Sharma 2 years, 11 months ago
Posted by Mukti Bhatti 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
The boiling point of the solution is 273.41 k
GIVEN
Weight of urea = 6 grams
weight of Glucose = 9 grams.
Weight of water = 300 grams
TO FIND
The boiling point of the solution.
SoOLUTION
We can simply solve the given problem as follows-
To calculate the boiling point of the solution, we will apply the following formula -
∆Tₒ = Kb× b × i (eq 1)
Where,
∆Tₒ = boiling point elevation
Kb = ebullioscopic constant of the solvent ( kb = o.515 kg/mol)
b = molality of the solute
i = van't Hoff factor of solute ( since urea and glucose are non-ionic compounds, so, i= 1)
We know that,
Boiling point elevation is defined as :
ΔTₒ = T₁ - T ........(eq2)
where,
T₁ = boiling point of the solution
T = boiling point of the solvent (T = 273.15 k of water)
Now,
{tex}\begin{aligned} & \text { number of moles of urea }=\frac{\text { given weight of urea }}{\text { molar mass of urea }} \\ & = \\ & \frac{6}{60}=0.1 \text { moles } \\ & \text { number of moles of glucose }=\frac{\text { given weight of glucose }}{\text { molar mass of glucose }} \\ & =\frac{9}{180}=0.05 \text { moles } \\ & \text { Total moles of solute }=0.1+0.05=0.15 \text { moles } \\ & \text { molality of solute }(b)=\frac{\text { moles of solute }}{\text { mass of solvent in } \mathrm{kg}} \\ & b=\frac{0.15 \times 1000}{300} \\ & b=0.5 \mathrm{M} \end{aligned}{/tex}
putting the values in ( eq 1) we have
ΔTₒ = 0.512 × 0.5 × 1
ΔTₒ = 0.26 k
Now, putting the value of ΔTₒ in (eq 2)
ΔTₒ = T₁ - T
0.26 = T₁ - 273.15
T₁ = 273.15 + 0.26
T₁ = 273.41 k
Hence, The boiling point of the solution is 273.41 k
Posted by Neeraj Yadav 2 years, 11 months ago
- 0 answers
Posted by Anoubam Varun 2 years, 11 months ago
- 1 answers
Posted by Manas Vichare 2 years, 11 months ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Alok Mittal 2 years, 9 months ago
0Thank You