Question

PLEASE HELP!!!

What is the molarity of the following substances:

a. 35.0g NaOH in 100ml H2O

b. 20.0g CuCl2 in 350ml H2O

c. 3.35g CaCO3 in 500ml H2O

Thanks in advance :)

Answer 1

a. 8.75 M NaOH

b. 0.425 M CuCl₂ or 0.43 M CuCl₂

c. 0.067 M CaCO₃ or 0.07 M CaCO₃

Explanation:

Molality is computed using the formula:

`M = (Moles\:of\:solute)/(Liters\:of\:solution)`

So first thing you need to do is determine how many moles of solute there are and divide it by the solution in liters.

Converting mass to moles, you need to get the mass of each solute per mole. You can use the periodic table to get the atomic mass (which is the grams per mole of each atom) of each of the elements involved. Then add them up and you will have how many grams per mole of each compound.

1. 35.0g of NaOH in 100ml H₂O

Element number of atoms atomic mass TOTAL

Na 1 x 22.99g/mol = 22.99g/mol

O 1 x 16.00g/mol = 16.00g/mol

H 1 x 1.01g/mol = 1.01g/mol

40.00g/mol

This means that the molecular mass of NaOH is 40.00 g/mol

Then we use this to convert 35.0g of NaOH to moles:

`35.0g \:of\:NaOH * (1\:mole\:of\:NaOH)/(40.00g\:of\:NaOH) = (35.0\:moles\:of\:NaOH)/(40.00)=0.875\:moles\:of\:NaOH`

Now that you have the number of moles we divide it by the solution in liters. Before we can do that you have to conver 100ml to L.

`100ml*(1L)/(1000ml) = 0.1 L`

Then we divide it:

`(0.875\:moles\:of\:NaOH)/(0.1L of solution) = 8.75M\: NaOH`

2. 20.0g CuCl₂ in 350ml H₂O

Element number of atoms atomic mass TOTAL

Cu 1 x 63.55g/mol = 63.55g/mol

Cl 2 x 34.45g/mol = 70.90g/mol

134.45g/mol

`20.g\:of\:CuCl_2*(1\:mole\:of\:CuCl_2)/(134.45\:g\:of\:CuCl_2)=0.1488\:moles\:of\:CuCl_2`

350ml = 0.350L

`(0.1488\:moles\:of\:CuCl_2)/(0.350L\:of\:solution)=0.425M\:CuCl_2`

3. 3.35g CaCO₃ in 500ml

Element number of atoms atomic mass TOTAL

Ca 1 x 40.08g/mol = 40.08g/mol

C 1 x 12.01g/mol = 12.01g/mol

O 3 x 16.00g/mol = 48.00g/mol

100.09g/mol

`3.35g\:of\:CaCO_3*(1\:mole\:of\:CaCO_3)/(100.09\:g\:of\:CaCO_3)=0.0335\:moles\:of\:CaCO_3`

500ml = 0.5L

`(0.0335\:moles\:of\:CaCO_3)/(0.5L)=0.067M\:CaCo_3`