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# Define molality. How many grams of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$ should be dissolved in ${\rm{250}}\,{\rm{grams}}$ of water to prepare ${\rm{0}}{\rm{.1m}}$ solution?

Last updated date: 24th Jun 2024
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Hint:
As we know that, the solution is made by dissolving a small number of particles to more number of solid or liquid. The small number of particles is known as solute particles and large numbers of particles are known as solvent. There are many ways to express concentration.

Complete step by step solution
A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

The concentration of a solution is the amount of the solute dissolved in the known amount of the solvent or solution. So, molality is the way to define concentration of a homogeneous solution.

We know that the concentration of the solution can be measured in terms of molarity or molality. The number of moles of the component can be expressed in terms of molality and molarity. Let’s consider them in detail.

The number of moles present in the solution can be calculated by dividing mass of substance with their molar mass. The number of moles of the component which is present in a small amount in a liter solution is termed as the molarity of the solution. The S.I. a unit of molarity is moles per liter. On the other hand, Molality of a solution is defined as the number of moles of the solute dissolved in ${\rm{1}}\,{\rm{kg}}$ of the solvent. It is denoted by ‘${\rm{m}}$’ and can be written as-
${\rm{m}} = \dfrac{{{\rm{number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solvent}}\,{\rm{in}}\,{\rm{g}}}} \times 1000 - - - - (i)$
We are given as,
Mass of solvent(water) is ${\rm{250}}\,{\rm{g}}$.
molality${\rm{(m)}} = {\rm{0}}{\rm{.1}}\,{\rm{m}}$
Molar mass of sodium carbonate is ${\rm{106}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$.
we have to find the mass of sodium carbonate (${\rm{w}}$).
putting given values in the above equation, we get as
$0.1 \, m \, = \dfrac{\rm{w/106} \, \dfrac{\rm{g}}{\rm{mol}}} {\rm{250 \, g}} \times 1000 \\$
By rearranging and calculating we get
$w \, = \, 2.65 g$

Note:
Molality is used for the solute dissolved in solvent and molarity is used for the solute dissolved in solution. Molarity is represented as ${\rm{(M)}}$.