Question

90 gm of glucose dissolve in 500 gm of water and density of the solution is 1.8g. find the:---
1) mole fraction of solute
2) %(w/w)
3) %(w/v)
4) molality


please answer this question correctly its very very urgent​

Answer 1

Answer :-

(1) Mole fraction of solute = 0.018

(2) % (w/w) = 15.25 %

(3) % (w/v) = 25.47 %

(4) Molality of the solution = 1 m

Explanation :-

We have :-

→ Mass of solute (glucose) = 90 g

→ Mass of solvent (water) = 500 g

→ Density of the solution = 1.8 g/mL

______________________________

Firstly, let's calculate the number of moles of solute (glucose) .

= Given Mass/Molar mass

= 90/180

= 0.5 mole

Now, number of moles of solvent (water) :-

= 500/18

= 27.78 moles

Mole fraction of solute

= Moles of solute/Total moles

= 0.5/(0.5 + 27.78)

= 0.5/28.28

= 0.018

______________________________

Mass of the solution :-

= Mass of solute + Mass of solvent

= (90 + 500) g

= 590 g

Volume of the solution :-

= Mass/Density

= 590/1.8

= 327.78 mL

Concentration of solution (w/w %)

= Mass of solute/Mass of solution × 100

= 90/590 × 100

= 0.1525 × 100

= 15.25 %

Concentration of solution (w/v %)

= Mass of solute/Volume of solution × 100

= 90/327.78 × 100

= 0.2745 × 100

= 27.45 %

Molality of the solution

= Moles of solute/Mass of solvent (kg)

= 0.5/(500 ÷ 1000)

= 0.5/0.5

= 1 m

Answer 2

Answer:

Question :-

• 90gm of glucose is dissolved in 500gm of water and density of solution is 1.8g.Find

1. mole fraction of solute
2. $( \frac{w}{w}) \%$
3. $\frac{w}{v} \%$
4. molarity.

Answers :-

1. mole fraction of solute =0.018
2. w/w% =15.254%
3. w/v% = 27.45744%
4. Molarity =1m.

Given :-

• 90 gm of glucose is dissolved in 500 gm of water and density of solution is 1.8g..

To find :-

1. mole fraction of solute
2. w/w%
3. w/v%
4. Molarity.

Solution :-

• Here refer the given attachment for better understanding.

Hope it helps u mate.

Thank you .