Question

1.82g of glucose(molar mass is 180) is dissolved in 25g of water. Calculate the molality and mole fraction of glucose.

Answer 1

Explanation:

(a) Molality :-

Molality = moles of solute (glucose)

mass of solvent (water)

Moles of glucose = C6H12O6

Molar mass = (12×6)+(1×2)+(16×6)

= 72+12+96

= 180

Mass of glucose = 1.82g

Therefore,,

moles = 1.82

180

= 0.01 moles

Therefore,,,it's molality =

Moles of glucose (0.01)

Mass of water(0.025)

= 0.01

0.025 molal

◾Molality = 0.4 molal

(b) Mole fraction :-

Mole of glucose = 0.01

Mole of water = 25 mass

molar mass

mass = 25g

molar mass H2O

= (1×2) +16

= 2+16

= 18

Moles = 25 = 1.388

18

Moles of glucose = 0.01

moles of water = 1.388

Mole fraction of glucose = mole of glucose

= 0.01

= 7.2×10³

Mole of water = 1.388

therefore,,

Mole fraction of glucose =

72 = 9

1000 1250

Answer 2

Answer:

Molality = 0.4 m

Mole fraction=  0.007

Explanation:

Given:

Mass of glucose = 1.82 g

Molar mass of glucose = 180 g/mol

Mass of water = 25 g.

1. Molality:

Molality of a solution is defined as the moles of solute per mass of solvent in kg.

$Molality =\frac{Moles\ of\ solute}{Mass\ of solvent\ in\ kg}$

Now, for the given solution, first calculate the moles of glucose.

$Moles\ of\ glucose\ = \frac{Mass\ of\ glucose}{Molar\ mass\ of\ glucose} =\frac{1.82\ g}{180\ g/mol}\\Moles\ of\ glucose\ =0.01\ moles$

Mass of solvent (water) = 25 g = 0.025 kg.

Therefore, the molality is determined as-

$Molality = \frac{0.01\ moles}{0.025\ kg}= 0.4 m$

Hence, the molality of glucose solution = 0.4 m

2. Mole fraction:

Mole fraction of glucose is determined as-

$Mole\ fraction =\frac{Moles\ of\ glucose}{Moles\ of\ glucose + Moles\ of\ water}$

Moles of glucose = 0.01 moles

Similarly, the moles of water are calculated as-

Mass of water = 25 g

Molar mass of water = 18.0 g/mol

$Moles\ of \ water\ = \frac{Mass\ of\ water}{Molar\ mass\ of\ water} =\frac{25 g}{18.0\ g/mol}\\ Moles\ of\ water\ = 1.39\ moles$

Thus, mole fraction is-

$Mole\ fraction\ of\ glucose\ =\frac{0.01\ moles}{0.01 + 1.39 } = 0.007$

Hence, mole fraction of glucose solution = 0.007.

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