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gtnhenbr [62]
4 years ago
9

A 5 grams sample of liquid water evaporates into gaseous water in a closed container, expanding in the process. What is the mass

after evaporation?
Chemistry
1 answer:
padilas [110]4 years ago
5 0

We have get the mass of gaseous water after evaporation in a closed container.

The mass of water vapor after evaporation is 5 grams.

In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.

Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.

Randomness of  molecules increases in gaseous state than liquid state.  


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What is the mass of KOH found in a 785 mL of a 1.43 M solution of KOH
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Molarity is defined as the ratio of number of moles to the volume of solution in litres.

The mathematical expression is given as:

Molarity = \frac{Number of moles}{volume of solution in litres}

Here, molarity is equal to 1.43 M and volume is equal to 785 mL.

Convert mL into L

As, 1 mL = 0.001 L

Thus, volume = 785\times 0.001 = 0.785 L

Rearrange the formula of molarity in terms of number of moles:

Number of moles =molarity\times volume of solution in litres

n = 1.43 M \times 0.785 L

= 1.12255 mole

Now,  Number of moles  = \frac{mass in g}{molar mass}

Molar mass of potassium hydroxide  = 56.10 g/mol

1.12255 mole  = \frac{mass in g}{56.10 g/mol}

mass in g = 1.12255 mole\times 56.10 g/mol

= 62.97 g

Hence, mass of KOH = 62.97 g



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