
Calculate the mass percentage of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ that is made by ${\text{10g}}$${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ in ${\text{70g}}$ water.
A.12.5
B.10
C.18
D.15
Answer
558.6k+ views
Hint: To solve this question, we must follow this step: firstly we calculate the mass of the solution. Mole is a physical quantity that represents the amount of mass of the substance required to have a collective of ${\text{6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}}$ atoms of the given substance.
Formula used
Mass percentage = $\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solution}}}}{{ \times 100}}$
Complete step by step answer:
Before we move forward with the solution of this question, let us understand some important basic concepts.
Mole: a mole is a physical quantity that represents the amount of mass of the substance required to have a collective of ${\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}$ atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance.
Mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ = ${\text{10g}}$(given)
Mass of water = ${\text{70g}}$(given)
Now we calculate the mass of the solution
To calculate the mass of the solution, we must calculate the mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and mass of water.
$\therefore $ Mass of the solution = Mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$+Mass of water
=${\text{10g}}$+${\text{70g}}$
= ${\text{80g}}$
The Mass percentage of the ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ can be calculated by applying this formula
Mass percentage = $\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solution}}}}{{ \times 100}}$
Mass of solute (${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$) = ${\text{10g}}$
Mass of the solution = ${\text{80g}}$
$\therefore $ Mass percentage = $\dfrac{{10}}{{80}}{{ \times 100 = 12}}{\text{.5 }}$
Thus Option A is correct.
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.
Formula used
Mass percentage = $\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solution}}}}{{ \times 100}}$
Complete step by step answer:
Before we move forward with the solution of this question, let us understand some important basic concepts.
Mole: a mole is a physical quantity that represents the amount of mass of the substance required to have a collective of ${\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}$ atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance.
Mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ = ${\text{10g}}$(given)
Mass of water = ${\text{70g}}$(given)
Now we calculate the mass of the solution
To calculate the mass of the solution, we must calculate the mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and mass of water.
$\therefore $ Mass of the solution = Mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$+Mass of water
=${\text{10g}}$+${\text{70g}}$
= ${\text{80g}}$
The Mass percentage of the ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ can be calculated by applying this formula
Mass percentage = $\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solution}}}}{{ \times 100}}$
Mass of solute (${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$) = ${\text{10g}}$
Mass of the solution = ${\text{80g}}$
$\therefore $ Mass percentage = $\dfrac{{10}}{{80}}{{ \times 100 = 12}}{\text{.5 }}$
Thus Option A is correct.
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.
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