What mass of glucose (C6H12O6) is needed to prepare 0.225m C6H12O6 using 250.g g of water?
The molality of a solution is defined as the amount of substance (in mol) of solute dissolved in solution is divided by the mass (in kg) of the solvent.
Molality(M): Number of moles of solute / Weight of solvent in kg
Here, 0.225m = x / 0.250Kg, where x = Number of moles of glucose.
So, glucose = 0.05625 moles.
Molar mass of glucose is 180.15 g/mol
So, the amoung of glucose will be = 0.05625 moles. X 180.15 g/mol - 10.13g
What mass of glucose (C6H12O6) is needed to prepare 0.225m C6H12O6 using 250.g g of water?
Solution:- molality(m) = moles of glucose/mass of water in kg
or, 0.225 = moles of glucose/0.25
or, moles of glucose = 0.05625
Now, molar mass of glucose = 180 g/mole
hence mass of glucose required = moles*molar mass = 10.125 g
Given: Molality of glucose ( C6H12O6 ) = 0.225m
==> Mass of solvent (water) = 250g
==> We know that Molality of glucose = ( Moles of glucose x 1000 ) / Mass of water
==> 0.225 = ( n x 1000 )/250
==> n = ( 0.225 x 250 ) / 1000
==> n = 0.225/4
==> Mass of glucose = No. of moles x Molar Mass of glucose
==> (0.225 x 180)/4
==> 10.125 grams ..........ANS
What mass of glucose (C6H12O6) is needed to prepare 0.225m C6H12O6 using 250.g g of water?
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