Question

Given, 10 mL of 6% (w/v) aqueous urea solution is diluted such that its molarity becomes 0.1 M. Volume of water added to make such solution is:
(a)- 70 mL
(b)- 80 mL
(c)- 90 mL
(d)- 990 mL

Answer 1

The formula that can be used for this question is ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$, where M is the molarity of the solutions and V are the different volumes. Convert the 6% (w/v) into grams and volume and then find the molarity.

Complete answer:
The given information in the question is 6% (w/v) urea solution. This means that the solute is in grams and the volume is in mL or L. 6% (w/v) means the amount of the urea is 6 grams and the volume of the solution is 100 mL.
This can also be written as 1000 mL or 1 L of the solution will have 60 grams of urea.
We know that the molecular mass of urea is $60\text{ g/mol}$
With this data, we can calculate the number of moles of the solution by dividing the given mass by the molecular mass.
$moles=\dfrac{60}{60}=1$
The mole in the solution is 1 and the volume is also 1 L, so the molarity will also be 1.
Now, this solution is diluted to such extent that the molarity becomes 0.1 M. So we can apply the formula ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$.
$1\text{ x 10 = 0}\text{.1 x }{{\text{V}}_{2}}$
${{V}_{1}}=10\text{ mL}$ (Given in the question)
${{V}_{2}}=\dfrac{10}{0.1}=100\text{ mL}$
Now the volume added to the solution will be: 100 – 10 = 90 mL.

So the correct answer is an option (c)- 90 mL.

Note:
The equation ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ is known as molarity equation is used when we have to compare two molarities of the solution. Similarly, we have normality equation ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$, which is used to compare two normalities of the solution.