Question

What is the molality of a solution made by dissolving 14.7 g of c6h12o6 into 150.0 ml of water? assume the density of water is 1.00 g/ml?

Answer 1

Hello!

We have the following data:

W (molality) = ? (in Molal)
m1 (mass of the solute) = 14.7 g
m2 (mass of the solvent) =? (in Kg)
d (solvent density) = 1,00 g/mL
V (volume of the solvent) = 150 mL

The mass of the solvent will be found by the following formula


`d = (m_2)/(V)`


`1 = (m_2)/(150)`


`m_2 = 1*150`


`m_2 = 150\:g\to \boxed{m_2 = 0,15\:Kg}`

M1 (Molar mass of solute - C6H12O6) = ?
C = 6*12 = 72 u
H = 12*81 = 12 u
O = 6*16 = 96 u
------------------------
M1 (Molar mass of solute - C6H12O6) = 72+12+96 = 180 g/mol

Now, let's find Molality, applying the above data to the formula, let's see:


`\omega = (m_1)/(M_1*m_2)`


`\omega = (14,7)/(180*0,15)`


`\omega = (14,7)/(27)`


`\boxed{\boxed{\omega \approx 0,54\:Molal}}\end{array}}\qquad\checkmark`

I hope this helps. =)