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3 years ago

Which of the compounds h2c2o4, ca(oh2, koh, and hi, behave as bases when they are dissolved in water?

2 answers:

6 0

H2C2O4, also written as (COOH)2, is Ethandioic acid, and, as evident by the name, is an acid, and not a base
Ca(OH)2 and KOH, Calcium Hydroxide and Potassium Hydroxide, respectively, are bases when dissolved in water, as they are able to release OH- ions.
Hydrogen Iodide, HI, when dissolved in water, is a strong acid. So, that is also not a base.

tatyana61 [14]3 years ago

5 0

Answer: The correct answer is $Ca(OH)_2\text{ and }KOH$

Explanation:

Acids are defined as the chemical species that produces hydrogen ions on dissolving in water.

Bases are defined as the chemical species that produces hydroxide ions on dissolving in water.

We are given:

Chemical compounds having chemical formulas $CH_3COOH,Ca(OH)_2,KOH\text{ and }HI$

$CH_3COOH$ is an organic acid and it produces $CH_3COO^-\text{ and }H^+$ ions

$Ca(OH)_2$ is a base and it produces $Ca^{2+}\text{ and }OH^-$ ions

$KOH$ is a base and it produces $K^{+}\text{ and }OH^-$ ions

HI is an acid because it produces $H^+\text{ and }I^-$ ions

Hence, the correct answer is $Ca(OH)_2\text{ and }KOH$

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

Answer: The freezing point of solution is -0.974°C

Explanation:

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

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

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

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

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

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

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

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$