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21- 36 g of glucose (molecular mass 180) is present in 500 g of water, the molality of the solution is

21- 36 g of glucose (molecular mass 180) is present in 500 g of water, the molality of the solution is

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Choose the correct answer and give the solution two points, please
21- 36 g of glucose (molecular mass 180) is present in 500 g of water, the molality of the solution is * O (a) 0.2 O (b) 0.4 O (c) 0. 8 O (d) 1.0 28- A mixture of three gases 02, N2 and CO2 is * O (a) 1-phase system O (b) 2-phase system O (c) 3-phase system O (d) 4-phase system
Transcribed Image Text:21- 36 g of glucose (molecular mass 180) is present in 500 g of water, the molality of the solution is * O (a) 0.2 O (b) 0.4 O (c) 0. 8 O (d) 1.0 28- A mixture of three gases 02, N2 and CO2 is * O (a) 1-phase system O (b) 2-phase system O (c) 3-phase system O (d) 4-phase system
Expert Solution
Step 1 molality

 

 

As per guidelines, if multiple questions are present , then first question is allowed to answer please post question seperately thank you 

 

Molality (m) :

Number of moles of solute is present in 1 Kg of solvent is called molality (m) 

m = nB / WA (in Kg)  

where 

nB (moles of solute)= given mass (W) / molecular mass (M) 

therefore

m = W/M x 1000/WA (in g)    --- (1) 

 

Units for molality (m) :  mol/Kg

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