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3 years ago

A 32 g sample of LiCl is dissolved in water to for 655 mL of solution. What is the molarity of the solution?

Chemistry

1 answer:

emmainna [20.7K]3 years ago

3 0

Answer : The molarity of the solution is, 1.1 M LiCl

Explanation : Given,

Mass of $LiCl$ = 32 g

Volume of solution = 655 mL

Molar mass of $LiCl$ = 42.4 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }LiCl\times 1000}{\text{Molar mass of }LiCl\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{32g\times 1000}{42.4g/mole\times 655mL}=1.1mole/L=1.1M$

Therefore, the molarity of the solution is, 1.1 M LiCl

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(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

Explanation:

Helium gas at temperature T=293k

Helium gas at pressure P= 500kPa

The inner diameter of spherical tank is $D_1 = 2m$

The inner radius of spherical tank is : $r_1 = \frac{D_1}{2}$

= $\frac{2}{2}$

=1m

Thickness of the container r = 1cm =0.01m

Outer radius of the spherical tank is ;

t = $r_2 - r_1$

$-r_2 = -t - r-1$

multiplying through with (-) we have ;

$r_2 = t + r_1$

$r_2 = 1 + 0.01m$

$r_2 = 1.01m$

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as

$D_A_B =4.0 ×10^-14 \frac{m^2}{s}$

From table molar mass and gas constant, the molecular weight of helium is:

M = 4.003kg/kmol

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.

$S_He$ = 0.00045 kmol/m³. bar

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as

$C_B_I = S_H_e×P$

= 0.00045kmol/m³. bar × (5)

$C_B_I = 2.25×10^-3 kmol/m³$

From one dimensional mass transfer through spherical layers is expressed as:

$N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}$

substituting all the values in the above relation, we have;

$M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}$

$M_di_f_f=11.42×10^-14kmol/s$

(a) The mass flow rate is expressed as

$M_di_f_f = MN_diff$

$M_di_f_f=4.003×11.42×10^-14$

$M_di_f_f=45.71×10^-14kg/s$

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

$N_di_ff =11.42×10^-14kmol/s$

$N_di_ff= 11.42×10^-14×7×24×3600 kmol/week$

$N_di_f_f=6.9×10^-8kmol/week$

The volume of the spherical tank is $V=\frac{4}{3} πr_1^3$

$V=\frac{4}{3}π×1^3$

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

$N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week$

$N_di_f_f-final ≅ 0.86$

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

3 0

3 years ago