Question

How many grams of NaCl are required to make 150.0 mL of a 5.000 m solution

Answer 1

Answer : The mass of NaCl required is 43.83 g.

Solution : Given,

Volume of solution = 150 ml

Molarity of solution = 5 mole/L

Molar mass of NaCl = 58.44 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of a solution.

Formula used :

`Molarity=(w_(solute)* 1000)/(M_(solute)* V_(solution(L)))`

where,

`{w_(solute)` = mass of solute

`{M_(solute)` = molar mass of solute

`{V_(solution(L))` = volume of solution in liter

In this question, the solute is NaCl.

Now put all the given values in this formula, we get

`5mole/L=(w_(solute)* 1000)/(58.44g/mole* 150L)`

By rearranging the term, we get the mass of NaCl.

`w_(solute)=43.83g`

Therefore, the mass of solute (NaCl) required is 43.83 g.

Answer 2

The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g

calculation

Step 1:calculate the number of moles

moles = molarity x volume in L

volume = 150 ml / 1000 = 0.15 L

= 0.15 L x 5.000 M = 0.75 moles

Step 2: calculate mass

mass = moles x molar mass

molar mass of NaCl = 23 + 35.5 = 58.5 mol /L

mass is therefore =0.75 moles x 58.5 mol /l =43.875 g