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 :

$Molarity=\frac{w_{solute}\times 1000}{M_{solute}\times V_{solution(L)}}$

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

$5mole/L=\frac{w_{solute}\times 1000}{58.44g/mole\times 150L}$

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