The solubility of KCl in water is 34.2 g per 100 g of water.
That means at STP, 34.2 g of KCl is dissolved in 100 g of water.
So, 100 g of water dissolved= 34.2 g of KCl
1 g of water dissolved=\frac{34.2 g of KCl}{100}
200 g of water dissolved=\frac{34.2 g of KCl}{100}\times 200
=68.4 g of KCl
So, the maximum amount of KCl that can dissolve in 200 g of water is 68.4 g.
Answer: You meant in constant temperature?
Because temperature is not mentioned at the beginning.
I suppose temperature remains constant, otherwise
there is no enough data for solving the problem
Explanation: According Boyle´s law in constant temperature
pV = constant. So p1V1 = p2V2 and pressure p2 = p1V1 /V2
= 88 atm · 18 l / 12 l = 132 atm.
Answer:
This is a chemical change because new substances are formed with different properties and identities.
The molar mass is 12.011 g/mol
