A solution is `0.150` mole fraction glucose `(C_(6)H_(12)O_(6))` and `0.850` mole fraction water `(H_(2)O)`. The molality of glucose in the solution is

Options

A. `9.80 m`
B. `8.90 m`
C. `7.86 m`
D. `6.87 m`

Answer

1

Explanation

Glucose is a sugar that occurs in fruits. It is also known as "blood sugar" because it is found in blood and is the body's main sources of energy. If we consider one mole of solution, then it contains `0.150 mol` of glucose and `0.850 mol` of water. We can convert moles of solvent `(H_(2)O)` to kilograms, to calculate the molarity.
Mass of `H_(2)O=(n_(H_(2)O))(molar mass_(H_(2)O))`
`=(0.850 mol)(18.0 g mol^(-1))`
`=15.3 g H_(2)O`
Molality of glucose `(C_(6)H_(12)O_(6))` in the solution is
`m=n_(glucose)/(g_(H_(2)O)) xx (1000 g)/(kg)`
`=(0.150 mol C_(6)H_(12)O_(6))/(15.3 g H_(2)O)xx(1000 g)/(kg)`
`=9.80 mol kg^(-1)`