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melamori03 [73]
3 years ago
15

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

concentrated sulfuric acid if the density of the solution is 1.83 g cm-3.
Chemistry
1 answer:
timama [110]3 years ago
7 0

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

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3 years ago
A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene
muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

\Delta T_f = K_f \times m

Where,

\Delta T_f = Depression in freezing point

K_f = Molal depression constant

m = Molality

\Delta T_f = K_f \times m

1.33 = 5.12 \times m

m = 0.26

Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}

Mass of solvent (toluene) = 15.0 g = 0.015 kg

0.26 = \frac{Mole\ of\ compound}{0.015}

Moles of compound = 0.015 × 0.26 = 0.00389 mol

Mol = \frac{Mass\ in\ g}{Molecular\ weight}

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Molecular\ weight = \frac{Mass\ in\ g}{Moles}

Molecular weight = \frac{1.450}{0.00389} = 372.13\ g/mol

4 0
3 years ago
Calculate the solubility of BaCO3 (a) in pure water and (b) in a solution in which [CO32-] = 0.289 M. Solubility in pure water =
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Answer:

(a). The solubility of BaCO_{3} in pure water is 4.4\times10^{-5}\ M

(b). The solubility of BaCO_{3} in solution is 6.92\times10^{-9}\ M

Explanation:

Given that,

(a). The solubility of BaCO_{3} in pure water

(b). The solubility of BaCO_{3} in a solution

Solubility of CO_{3}^{-2} is 0.289 M

We know that,

The solubility product constant  of  BaCO_{3} is 2\times10^{-9}

Let the solubility of BaCO_{3} is s.

We need to calculate the solubility of BaCO_{3} in pure water

Using formula of solubility

ksp=s\times s

ksp=s^2

s=\sqrt{ksp}

Put the value into the formula

s=\sqrt{2\times10^{-9}}

s=4.4\times10^{-5}\ M

(b). We need to calculate the solubility of BaCO_{3} in solution

Using formula of solubility

ksp=s\times s

Put the value into the formula

2\times10^{-9}=s\times 0.289

s=\dfrac{2\times10^{-9}}{0.289}

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(b). The solubility of BaCO_{3} in solution is 6.92\times10^{-9}\ M

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Answer:

See the answer below

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The interphase is divided into;

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A cell at the end of m phase undergoes massive growth and development at the G1 phase, doubles its DNA at the S phase through replication, and synthesizes proteins at the G2 phase before hopping into the m phase again.

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