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GenaCL600 [577]
3 years ago
15

Calculate the molar concentration of HA (90.2 g/mol) in a solution that has a specific gravity of 1.24 and is 64% HNO3 (w/w).​

Chemistry
1 answer:
ser-zykov [4K]3 years ago
8 0

Answer:

Explanation:

Hello,

In this case, considering the given specific gravity we are to compute the density of the 64% HNO3 solution:

\rho=1.24*1g/mL=1.24g/mL

In such a way, since the given solution of nitric acid is 64 % by mass, we've got:

\frac{64gHNO_3}{100g\ solution}

Thus, by using the previously computed density and the molar mass of HA (90.2 g/mol) we can compute the required molar concentration as follows:

M=\frac{(100g-64g)HA}{100g\ solution}*\frac{1.24g\ solution }{1mL \ solution}*\frac{1000mL\ solution}{1L\ solution}*\frac{1molHA}{90.2gHA}   \\\\M=4.95M

Regards.

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The osmotic pressure of an aqueoussolution of urea at 300 K is
den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=imRT

where,

\pi = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the solution = 300 K

Putting values in above equation, we get:

120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=i\times K_f\times m

where,

i = Vant hoff factor = 1 (for non-electrolytes)

K_f = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.  

  • The equation used to calculate freezing point of solution is:

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

where,

\Delta T = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC

Hence, the freezing point of solution is -0.09°C

4 0
3 years ago
A key step in the extraction of iron from its ore is FeO(s) + CO(g) ⇋ Fe(s) + CO2(g) Kp =0.403 at 1000°C. What are the equilibri
andriy [413]

Answer:

0.713atm for CO and 0.287atm for CO₂

Explanation:

Based on the reaction:

FeO(s) + CO(g) ⇋ Fe(s) + CO₂(g)

Kp is defined as:

Kp = \frac{P_{CO_2}}{P_{CO}} = 0.403

When 1.00 atm of CO react with an excess of FeO, the pressures in equilibrium are:

PCO = 1.00atm - x

PCO₂ = x

<em>Where x represents the reaction coordinate.</em>

Replacing in Kp expression:

0.403 = \frac{x}{1-x}

0.403 - 0.403x = x

0.403 = 1.403x

0.287atm = x

Thus, pressures in equilibrium are:

PCO = 1.00atm - x = <em>0.713atm</em>

PCO₂ = x = <em>0.287atm</em>

<em></em>

8 0
3 years ago
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