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3 years ago

What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?

Chemistry

1 answer:

trapecia [35]3 years ago

7 0

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution = 2.5L

Mass of Calcium phosphate = 600g

Unknown:

Concentration = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

Formula of calcium phosphate = Ca₃PO₄

molar mass = 3(40) + 31 + 4(16) = 215g/mol

Number of moles of Ca₃PO₄ = $\frac{600}{215}$ = 2.79moles

Now;

Concentration = $\frac{Number of moles }{volume }$

Concentration = $\frac{2.79}{2.5}$ = 1.12M

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3 years ago

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2 years ago

The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

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$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

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Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

5 0

3 years ago

What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​

What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?