The molality of a solution containing 36g of glucose (C6H12O6) in 250g of water is:
(a) 0.4 m
(b) 0.6 m
(c) 0.8 m
(d) 1.2 m
Answer (2)
The molality of a solution containing 36 g of glucose in 250 g of water is approximately 0.8 m. This was calculated by determining the number of moles of glucose and the mass of water in kilograms. Thus, the chosen option is (c) 0.8 m.
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To find the molality of the solution, we first need to understand the concept. Molality ( m ) is defined as the number of moles of solute per kilogram of solvent. It is expressed in mol/kg.
First, we'll calculate the number of moles of glucose. The molecular formula of glucose is C 6 H 12 O 6 , and its molar mass is calculated as follows:
Carbon (C): 12 g/mol × 6 = 72 g/mol
Hydrogen (H): 1 g/mol × 12 = 12 g/mol
Oxygen (O): 16 g/mol × 6 = 96 g/mol
Adding these together gives a molar mass of glucose of 180 g/mol .
Next, calculate the moles of glucose:
Moles of glucose = 180 g/mol 36 g = 0.2 moles
The mass of the solvent (water) is given as 250 g , which is equivalent to 0.250 kg .
Now, calculate the molality using the formula for molality:
m = kilograms of solvent moles of solute = 0.250 kg 0.2 moles = 0.8 m
Therefore, the molality of the solution is 0.8 m , which corresponds to option (c) 0.8 m .
