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Mila [183]
3 years ago
7

Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

Chemistry
1 answer:
marin [14]3 years ago
5 0

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

1 mL = 0.001 L

Molarity = \frac{n}{V(L)}

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

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Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
pV=nRT
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}

If the supported load is W kg, then
p_{1} =  \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} =  20.3718W \, Pa
Similarly,
p_{2} =  \frac{4W}{\pi d_{2}^{2}} \, Pa

\frac{p_{1}}{p_{2}} =  \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}

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The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

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