Question

How many grams of sodium bicarbonate are needed to make 23.75ml of a 4.861 m solution

Answer 1

9.6936 grams of sodium bicarbonate are needed to make 23.75ml of a 4.861 M solution.

Step-by-step explanation:

Sodium bicarbonate is a compound of formula NaHCO₃.

Molarity is defined as the number of moles of solute (substance in minor proportion in a solution) in a given volume.

`Molarity=(number of moles)/(volume)`

Molarity is expressed in units (moles / liter).

So, 4,861 M means that in 1 L of solution there are 4,861 moles. Knowing that 23.75 mL equals 0.02375 L (1 L = 1000 mL), it is possible to apply a rule of three as follows: if in 1 liter of solution there are 4,861 moles of sodium bicarbonate, in 0.02375 L of solution, how many moles of the compound there is?

`moles=(0.02375L*4.861 moles)/(1 L)`

moles=0.1154

On the other side, the atomic mass of the elements that make up the sodium bicarbonate are:

• Na: 23 g/mol
• H: 1 g/mol
• C: 12 g/mol
• O: 16 g/mol

So, the mass of the sodium bicarbonate is:

NaHCO₃: 23 g/mol + 1 g/mol + 12 g/mol + 3*16 g/mol= 84 g/mol

Then, a rule of three can be applied to know the mass of sodium bicarbonate that are needed to make 23.75ml of a 4.861 M solution. The rule of three applies as follows: if, according to the molar mass of the sodium bicarbonate, in 1 mole there are 84 grams, in 0.1154 moles (previously calculated moles) how many grams are present?

`mass=(0.1154moles*84 grams)/(1 mole)`

mass=9.6936 grams

9.6936 grams of sodium bicarbonate are needed to make 23.75ml of a 4.861 M solution.

Answer 2

96.99 g NaHCO_3

Step 1. Calculate the moles of NaHCO_3

Moles of NaHCO_3 = 0.2375 L solution × (4.861 mol NaHCO_3/1 L solution)

= 1.1545 mol NaHCO_3

Step 2. Calculate the mass of NaHCO_3

Mass of NaHCO_3

= 1.1545 mol NaHCO_3 × (84.01 g NaHCO_3/1 mol NaHCO_3)

= 96.99 g NaHCO_3