Question

How many gram of KCl are present in 85,0ml of 2,10m kCl?

Answer 1

What is the molarity of a solution prepared by dissolving 5433g of ca(NO3)2 in 355ml of water?

Answer 2

To solve this problem, we can use the formula:

moles of solute = concentration x volume (in liters)

Then, we can convert moles to grams using the molar mass of KCl.

First, we need to convert the volume from milliliters to liters:

85.0 ml = 0.085 L

Next, we can plug in the given values:

moles of KCl = 2.10 mol/L x 0.085 L
moles of KCl = 0.1785 mol

Finally, we can convert moles of KCl to grams:

mass of KCl = moles of KCl x molar mass of KCl
mass of KCl = 0.1785 mol x 74.55 g/mol
mass of KCl = 13.3 g

Therefore, there are 13.3 grams of KCl present in 85.0 ml of 2.10 M KCl.

Answer 3

To calculate the molarity of the solution, we need to first calculate the number of moles of Ca(NO3)2 present in the solution.

Number of moles = mass / molar mass

The molar mass of Ca(NO3)2 is:

Ca = 1 x 40.08 = 40.08 g/mol
N = 2 x 14.01 = 28.02 g/mol
O = 6 x 16.00 = 96.00 g/mol

Molar mass of Ca(NO3)2 = 164.10 g/mol

So, the number of moles of Ca(NO3)2 in the solution is:

Number of moles = 5433 g / 164.10 g/mol = 33.10 mol

Now, we need to calculate the volume of the solution in liters:

Volume = 355 ml / 1000 ml/L = 0.355 L

Finally, we can calculate the molarity of the solution using the formula:

Molarity = number of moles / volume of solution in liters

Molarity = 33.10 mol / 0.355 L = 93.1 M

Therefore, the molarity of the solution prepared by dissolving 5433 g of Ca(NO3)2 in 355 ml of water is 93.1 M.