Question

A solution of barium hydroxide, $Ba{(OH)_2}$ contains $4.285 \cdot g$ of barium hydroxide in $100 \cdot mL$ of solution. What is the molarity of the solution?

Answer 1

Molarity is defined as the number of moles of component per liter of solution. It explains the strength of solution or we can say explains the concentration of solution. Barium hydroxide $Ba{(OH)_2}$ is a basic compound with molar mass of $171.34 \cdot g \cdot mo{l^{ - 1}}$ . First, the number of moles of barium hydroxide has to be calculated and then the volume is to be converted into liters.

Complete answer:
The molarity can be calculated as $molarity = moles/volume(liter)$
Now, the number of moles of barium hydroxide present in the given solution will be –
$moles = \dfrac {\text{mass}}{\text{molar mass}}$
So, the number of moles of $Ba{(OH)_2}$ will be $\dfrac{{4.285}}{{171.34}} = 0.02500 \cdot moles$
Now, volume in liters will be - $\dfrac{{100ml}}{{1000}} = 0.1liter$
Now, putting values in molarity formula, we have $0.02500mol \div 0.1l = 0.25mol/l$
Hence, the molarity of a given solution of barium hydroxide will be $0.25mol/l$ .

Note:
The conversion of mass into moles should be done carefully, the zeroes should be placed correctly. For converting mL into liters, the volume should be divided by $1000$ . Units should be written clearly. An alternate method uses the following formula if the mass is given in grams and volume is given in mL.
$molarity = \dfrac{{m(grams) \times 1000}}{{M(g \cdot mo{l^{ - 1}}) \times V(mL)}}$ , where m is mass given in grams, M is molar mass in gram per mole and V is volume in milliliter.