An

Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

8 0

3 years ago

¿Cuál era su juego favorito cuando eran niños?

blagie [28]

Answer:

mi juego favorito probablemente sería Mine craft cuando era más joven, aunque sigue siendo uno de mis juegos favoritos

Explanation:

7 0

2 years ago

Write Qc for each of the following:(1) Gaseous sulfur tetrafluoride reacts with liquid water to produce gaseous sulfur dioxide a

snow_tiger [21]

Answer : The expression for reaction quotient will be :

(1) $Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) $Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

Explanation :

Reaction quotient $(Q_c)$ : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

(1) The given balanced chemical reaction is,

$SF_4(g)+2H_2O(l)\rightarrow SO_2(g)+4HF(g)$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :

$Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) The given balanced chemical reaction is,

$2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

7 0

2 years ago

Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3

Vadim26 [7]

Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g

7 0

2 years ago

Read 2 more answers

What is the value of R in the ideal gas law?

dybincka [34]

It’s option B 0.0821L.atm/mol.K

5 0

2 years ago