Question

A sample of ${H_2}S{O_4}$ ​ (density =1.4 g/ml) is labelled as 49% by weight. What volume of acid has to be used to make 1.0 L of 0.2 M ${H_2}S{O_4}$​ solution?
A. 2000 ml
B. 28.57 ml
C. 13.6 L
D. 1.12 L

Answer 1

Hint: To get this problem solved we need to find the Molarity of ${H_2}S{O_4}$ with the help of given data then we have to apply the formula of dilution to get the volume of acid used. Doing this will solve your problem.

Complete answer:
Formula used - ${M_1}{V_1} = {M_2}{V_2}$.
Density of ${H_2}S{O_4}$ is given as 1.4g/ml
Then we can say 100ml of ${H_2}S{O_4}$ will have weight = 1400g
We can calculate the mass of ${H_2}S{O_4}$as $1400 \times \dfrac{{49}}{{100}}$​= 686g as ${H_2}S{O_4}$ is 49% by weight.
Then the number of moles of ${H_2}S{O_4}$ $\dfrac{{given\,weight}}{{molecular\,weight}} = \dfrac{{686}}{{98}} = 7$.
Therefore the molarity of the stock solution will be 7M.

To prepare 1000ml of 0.2M solution we need to use the formula of dilution:
${M_1}{V_1} = {M_2}{V_2}$
Dilution Formula-The concentration of the solution is most generally expressed in terms of mass level, mole fraction, molarity, molarity and normality. When calculating dilution factors, it is important that volume and concentration units remain consistent. Calculations of dilution can be made using the formula ​${M_1}{V_1} = {M_2}{V_2}$.
7 x ${V_1}$=0.2 x 1000
${V_1}$ = 28.57ml
Hence the needed volume is 28.57ml.
So, option B is correct.

Note: In order to solve such problems we need to understand that when we are getting liquid from the liquid itself but both the liquids are different in weight then we can use the formula of dilution. Dilution formula which is ${M_1}{V_1} = {M_2}{V_2}$ are very useful in such problems and it is not valid only for two sets of liquid but is valid for more sets if we are diluting the liquid at the same time. Knowing this will help you a lot and will get your problem right.