Question

How many milliliters of water are needed to make a 0.171 M solution that contains 1.00 g of NaCl?

Answer 1

Hint: Use the formula of molarity and find the volume, as all the other variables are given in the question. More the amount of salt that is added, more will be its molarity, if the volume of the water is kept constant.

Complete answer: In order to answer the question, we need to know about moles and molar mass. Now, matter is made up of atoms, and as matter has mass, then the atoms should have an individual mass. Molar mass of an element or compound is the mass which houses $6\times {{10}^{23}}$ particles. For, example, the hydrogen molecule has a molar mass of 2 grams. This means 2 grams of hydrogen contains $6\times {{10}^{23}}$atoms, and this number is also called the Avogadro’s number.
Number of moles of an element or a compound is the ratio of its given mass taken by the user, to its molar mass. More is the number of moles, more is the concentration of the substance. Now, let us come to the question. We first convert NaCl to the number of moles by using the formula $no\,of\,mole=\dfrac{given\,mass}{molar\,mass}$. So, we obtain the number of moles of NaCl as $\dfrac{1}{58.44}=0.0171mol$. Let us recall the formula for molarity which is:
\[Molarity=\dfrac{given\,mass}{molar\,mass}\times \dfrac{1000}{vol\,of\,sol}\]
Now, by using this formula, we will find the volume of solution required, as all the other data is given in the question. So, we have:
\[
   vol\,of\,sol=\dfrac{no\,of\,mole}{Molarity} \\
  \Rightarrow vol\,of\,sol=\dfrac{0.0171mol}{0.171mol{{L}^{-1}}}=0.100L \\
\]
So, we have 0.100L as the volume of solution, on converting it to millilitres, as asked in the question , we find the volume to be 100mL solution, which is the required answer.

Note: It is to be noted that volume of the solution can expand, when the solution is heated up, in that case, the molarity shall reduce, and vice versa. In this case, the molarity of the solution depends on the temperature, the volume of the solution, and the moles of the salt used.