A solution is 12.5% Na3PO, by mass. How many grams of Na3PO4
1 answer:
You might be interested in
Answer:
C）52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
Answer:
To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
Explanation:
Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.
So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles
We know, number of moles of a compound is the ratio of mass to molar mass of that compound.
So, mass of 2 moles of urea =
Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
So, option (C) is correct.