Question

What mass of sodium carbonate, $N{a_2}C{O_3}$ is needed to make 120 ml of a 1.5 M solution? (Given formula weight of sodium carbonate = 106 amu)
A)295 g
B)9.5 g
C)19 g
D)589 g
E)19000 g

Answer 1

To solve this question, we must follow these steps: first, we need to find the number of moles of $N{a_2}C{O_3}$ present in the solution. Then to find the mass of $N{a_2}C{O_3}$ , we must multiply the number of moles formed with the molar mass of $N{a_2}C{O_3}$ .

Formula used: Molarity $= \dfrac{{No.\,\,of\,\,moles}}{{volume}}$

Complete Step-by-Step Answer:
Before we move forward with the solution of this question, let us understand some important basic concepts.
1)Mole: a mole is a physical quantity which represents the amount of mass of the substance required to have a collective of $6.022 \times {10^{23}}$ atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.
Molarity: molarity can be understood as the physical quantity which represents the number of moles of a substance present in 1 litre of the solution. Mathematically, this relation can be represented as:
Molarity $= \dfrac{{No.\,\,of\,\,moles}}{{volume}}$
In this case, since volume of the solution is given, we can calculate the number of sodium bicarbonate using the concept of molarity. The equation given above can be rearranged as:
No. of moles = (volume) (molarity)
No. of moles $= \left( {0.120L} \right)\left( {1.5M} \right)$
No. of moles = 0.180 mol
Now, we have been given the data that the molecular weight of sodium bicarbonate is 106 $g/mol$ . Hence, the mass of Sodium Carbonate present in the give solution can be calculated as:
Mass of $N{a_2}C{O_3}$ = (molar mass) (no. of moles)
Mass of $N{a_2}C{O_3}$ $= \left( {106} \right)\left( {0.180} \right)$
Mass of $N{a_2}C{O_3}$ = 19 g
Hence, mass of sodium carbonate, $N{a_2}C{O_3}$ is needed to make 120 ml of a 1.5 M solution is 19 g.
Hence, Option C is the correct option.

Note: Mole concept simplifies the mass relation among reactants and products such that we can base our calculation on the coefficients (numbers of molecules involved in the reaction). At the same time, mass or the quantity of substance is on lab scale in grams.