What is the molarity and molality of a solution prepared by mixing 23g of cacl2 with 217g of water(assuming density of water is 1g/mol)

Molarity = 0.9 M

Molality = 0.95 m

Explanation:

Data given:

mass of CaCl₂ = 23 g

mass of water = 217g

Density of water = 1g/mol

molality of solution = ?

molarity of solution = ?

Solution:

Molarity and Molality are terms used for concenteration of solution

Formula Used for Molarity

                    Molarity = moles of solute / liter of solution . . . . . (1)

So first we find number of moles of 23 g CaCl₂.

Formula used to find moles

                no. of moles = mass in g / molar mass

molar mass CaCl₂ = 40 + 2(35.5)

molar mass CaCl₂ = 40 + 71 = 111 g/ mole

So,

Put values in mole formula

                  no. of moles = 23 g / 111 g/mol

                   no. of moles = 0.21 mol

Now,

As for molarity we have to convert grams of water to liter

for this purpose we will use density formula

                      d= m/v

for volume rearrange the above equation

                     v = m/d . . . . . . . (2)

put vlue in above equation 2

                  v = 217 g / 1 (g/mL)

                    v = 217 mL

Now,

Total amount of solution = 23 + 217 = 240 mL

Now, Convert the mL to Liter

1000 = 1 L

240 mL = 240 /1000 = 0.24 L

So no we have the following required values to find molarity

no. of moles of CaCl₂ = 0.21 mol

liter of solution = 0.24 L

Put values in equation 1

                 Molarity = 0.21 mol / 0.24 L

                 Molarity = 0.9 M

To find Molality (m)

Formula used to find Molality

                 Molality = moles of solute / kg of solvent . . . . . . . (3)

So, no we have to convert grams of water to Kg

1000 g = 1 kg

217 g = 217 / 1000 = 0.22 Kg

Now,

Put values in equation

                 Molality = 0.21 mol / 0.22 kg

                 Molality = 0.95 m

So, the Molality =  0.95 m