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4 years ago

What is the chemical formula for deionized water?

Chemistry

1 answer:

Shalnov [3]4 years ago

7 0

Answer:

Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)

Explanation:

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2 years ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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3 years ago