73.5 grams C, 4.20 grams H, and 72.3 grams Cl. What is the percent composition of each element in this compound?
1 answer:
You might be interested in
Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L
240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.
Explanation:
As per the solubility curve graph of different salts dissolved in 100 g of water, it can be noted that the KClO₃ gets saturated at 100 °C with the solubility of 59 g in it. But here the water is taken as 300 g and the salt is taken as 120 g. The graph shows the solubility of grams of salt in 100 g of water.
So, then we have to find what is the gram of salt present in 100 g of water for the present sample.
300 g of water contains 120 grams of salt.
Then, 1 g of water will contain g of salt.
Then, 100 g of water will contain of salt. This means, 100 g of water will contain 40 g of salt in it.
Then, if we check the graph, the corresponding x-axis point on the curve for the y-axis value of 40 g will give us the temperature required to dissolve the salt in water. So 80°C will be required to dissolve 40 g of salt in 100 g of water.
Then, in order to get the temperature required to dissolve 120 g in 300 g of water, multiply the temperature by 3, which will give the result as 80×3=240 °C.
So, 240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.