There is a solution of 50 gm Pb(NO3)2 & 75 ml boiled water.If the solution is taken to the temperature at 20°c then what quantity of Pb(NO3)2 will come out from the solution if the solubility of Pb(NO3)2 in water at 20°c is 54.
To determine the quantity of Pb(NO3)2 that will come out of the solution, we first need to calculate the maximum amount of Pb(NO3)2 that can dissolve in the given volume of water at 20°C. This maximum amount is called the solubility.
The solubility of Pb(NO3)2 in water at 20°C is given as 54 grams. This means that at 20°C, 54 grams of Pb(NO3)2 can dissolve in 100 ml of water.
To find the amount of Pb(NO3)2 that can dissolve in 75 ml of water, we will use the concept of proportion:
54 grams -> 100 ml
x grams -> 75 ml
To solve this proportion, we can use cross-multiplication:
54 grams × 75 ml = 100 ml × x grams
x = (54 grams × 75 ml) / 100 ml
Now, let's calculate the value of x:
x = (54 grams × 75 ml) / 100 ml
x = 40.5 grams
Therefore, at 20°C, the maximum amount of Pb(NO3)2 that can dissolve in 75 ml of water is 40.5 grams.
Since the initial solution contains 50 grams of Pb(NO3)2, which is more than the maximum solubility of 40.5 grams, we can conclude that 40.5 grams of Pb(NO3)2 will come out of the solution when it is cooled to 20°C.