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2 years ago

A sample of water contains 5.24 x 1022 molecules. How many moles are in this sample?

Chemistry

1 answer:

LekaFEV [45]2 years ago

6 0

Answer:

0.087 moles of water

Explanation:

Given data:

Number of molecules of water = 5.24×10²² molecules

Number of moles of water = ?

Solution:

Avogadro number:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules of water

5.24×10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.87×10⁻¹ mol

0.087 mol

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

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 = 2

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

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

2 years ago

Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys

8090 [49]

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

$Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}$ .

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both $[\mathrm{SO_2\, (g)}]$ and $[\mathrm{O_2\, (g)}]$ will increase if the pressure is increased through compression. However, because $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient $Q$ .

As a result, the increase in pressure will have no impact on the value of $Q\!$ . If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.