Question

A solution contains 48.6 g glucose (c6h12o6) dissolved in 0.800 l of water.what is the molality of the solution? (assume a density of 1.00 g/ml for water.)

Answer 1

`m=0.338mol/kg=0.338m`

Step-by-step explanation:

Hello,

In this case, since the molality is defined in terms of the moles of solute divided by the kilograms of solvent as shown below:

`m=(mol_(solute))/(V_(solution))`

One firstly computed the moles of glucose which acts as the solute as follows:

`mol_(solute)=48.6gC_6H_(12)O_6*(1molC_6H_(12)O_6)/(180gC_6H_(12)O_6) =0.27molC_6H_(12)O_6`

Now, as the density of water is 1.00 g/mL, we conclude we have 0.8 kg of water, therefore, the resulting molality in molar units (m) turns out:

`m=(0.27molC_6H_(12)O_6)/(0.8 kg)\\m=0.338mol/kg=0.338m`

Best regards.

Answer 2

Answer is: molality is 0,3375 mol/kg.
m(C₆H₁₂O₆) = 48,6 g.
V(H₂O) = 0,800 L · 1000 mL/1L = 800 mL.
d(H₂O) = 1 g/mL.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 800 mL · 1 g/mL.
m(H₂O) = 800 g ÷ 1000 g/1kg = 0,8 kg.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 48,6 g ÷ 180,15 g/mol.
n(C₆H₁₂O₆) = 0,27 mol.
b(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ m(H₂O).
b(C₆H₁₂O₆) = 0,27 mol ÷ 0,8 kg = 0,3375 mol/kg.