How much water is required to dissolve 1.00 g of PbCl2?
at 20C, the solubility is
Solubility in water 10.8 g/L (20 °C)
this would indicate 1 gram/(1000/10.8) = 1 gram/92.6ml
To determine the amount of water required to dissolve 1.00 g of PbCl2, we need to know the solubility of PbCl2 in water. The solubility of PbCl2 in water is 26.6 g/100 mL at 25°C.
To find the amount of water required, we can set up a simple proportion:
(solubility of PbCl2)/(volume of water) = (mass of PbCl2)/(mass of water)
Let's plug in the values:
(26.6 g/100 mL)/(x mL) = 1.00 g/(mass of water)
Simplifying the equation:
26.6/100 = 1.00/(mass of water)
Cross-multiplying:
26.6 * (mass of water) = 1.00 * 100
Dividing both sides by 26.6:
(mass of water) = (1.00 * 100) / 26.6
Calculating the result:
(mass of water) ≈ 3.76 g
Therefore, approximately 3.76 grams of water is required to dissolve 1.00 g of PbCl2.
To determine the amount of water required to dissolve 1.00 g of PbCl2, we need to consider the solubility of PbCl2 in water.
The solubility of a compound indicates the maximum amount of that compound that can dissolve in a given amount of solvent (in this case, water) at a specific temperature. To find the solubility of PbCl2, we can refer to a solubility table or database.
According to the solubility table, the solubility of PbCl2 at 25°C is 0.85 g/100 mL of water. This means that 0.85 grams of PbCl2 can dissolve in 100 mL (or 0.1 liters) of water.
To determine the volume of water required to dissolve 1.00 g of PbCl2, we can set up a proportion using the solubility information:
0.85 g of PbCl2 / 0.1 L of water = 1.00 g of PbCl2 / x L of water
Solving this proportion, we can find the volume of water required:
x L = (0.1 L * 1.00 g of PbCl2) / 0.85 g of PbCl2
x L = 0.118 L
Therefore, approximately 0.118 liters (or 118 mL) of water is required to dissolve 1.00 g of PbCl2 at 25°C.