Question

1.How many gram of concentrated sulphuric acid, H2SO4 are contained in 3.0 liters of 0.800 N solution

2. If the wine is 15 %, how many milliliters of ethanol, C2H5OH are present in a bottle containing 500 ml of wine?

3. What is the mole fraction of a solute in a 2.2 molal solution in which naphthalene (C10H8) is the solvent?

Atomic Masses (g/mole): Al=27 Na=23 O=16 H=1 C=12 S=32

Answer 1

Answer:

The percentage concentration of any solution is most commonly expressed as mass percent:

Mass % of any component of the solution =

(Mass of the component in the solution / Total mass of the solution) x 100

Other methods are:

Volume percentage:

Volume % of a component =

(Volume of the component/Total volume of the solution) x 100

Mass by volume percentage:

It is the mass of solute dissolved in 100 mL of the solution.

i.e. Mass by Volume percentage =

(Mass of solute in grams/Volume of solution in mL) x 100Explanation:

Answer 2

1. Moles of the given solution $=0.8N$

The volume of the solution given $=3.0L$

It is known to us that any solution of normality 1 will be equal to 1 gram equivalent of the solute dissolved in that solvent.

The equivalent weight of $H_2SO_4$ = $49gm$

If 1 gram equivalent of concentrated sulphuric acid $=49gm$

∴ $0.8N$ equivalent of concentrated sulphuric acid $=49*0.8=39.2gm$

∴ 39.2gm of concentrated sulphuric acid is contained in 3.0 litres of 0.800N solution.

2. Percentage of wine in the ethanol $=15$%

In this case, wine is solution and ethanol is solute,

The volume of the solution $=500ml$

We know,

Percentage by Volume = $\frac{Volume-of-solute}{Volume-of-the-solution}*100$

⇒ $15=\frac{Volume-of-solute}{500}*100$

⇒ $15*5$ = Volume of the solute

∴ The volume of solute, ethanol is 75mL.

3. Molality of the solution $=2.2M$

For naphthalene, the molar mass will be

$=C_1_0H_8=12*10+8*1=128g/mol$

According to the relation,

Molality = Mole fraction × $\frac{1000}{Molar-Mass}$

⇒ Mole fraction = Molality × $\frac{Molar-Mass}{1000}$

⇒ Mole fraction = 2.2 × $\frac{128}{1000}$

⇒ Mole fraction = 0.28

∴ The mole fraction of a solute will be 0.28.