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2 years ago

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Seawater has a salinity of 3.5%, i.e. 35 g of solid NaCl for one kilogram of seawater. For simplicity, assuming that all dissolv

ed salts in seawater are NaCl rather than Na and Cl- ions and other species, what is the osmotic pressure between seawater and fresh water at 300K?

1 answer:

artcher [175]2 years ago

6 0

Answer:

π = 29.3 atm

Explanation:

The osmotic pressure π is given by

π = i MRT, where i=Vant Hoff factor =2 for NaCl ( 2 moles ions/mol NacL)

M = molarity

R = 0.0821 Latm/molK

T = temperature K

Assuming density of solution is 1 g/mL

M= 35 g/58.44 gmol / 1 L = 0.60 M

π = 2.0 x 0.60 mol/L x 0.0821 Latm/kmol x 300K = 29.3 atm

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Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

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If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

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2 years ago

Calculate the molar concentration of the Br⁻ ions in 0.51 M MgBr2(aq), assuming that the dissolved substance dissociates complet

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MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
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The answer to this question is [Br⁻] = 1.0 M

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2 years ago

Which item has the most thermal energy?

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

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