Question

The density of $1M$ solution of $HCl$ is$1.0585g/mL$. The molality of the solution is.
A) $1.0585$
B) $1.00$
C) $0.10$
D) $0.0585$

Answer 1

Hint: We know that the molality of a solution is characterized as the measure of substance (in moles) of solute, no. solute, separated by the mass (in kg) of the dissolvable, m dissolvable.
${\text{Molality}} = \dfrac{{{\text{No}}{\text{. of solute}}}}{{{\text{Mass of dissolvable}}\left( {kg} \right)}}$
In the instances of arrangements with more than one dissolvable, molality can be characterized for the blended dissolvable considered as an unadulterated pseudo-dissolvable. Rather than mole solute per kilogram dissolvable as in the double case, units are characterized as mole solute per kilogram blended dissolvable.

Complete step by step answer:
We can calculate the molality using the equation is,
${\text{Molality}} = \dfrac{{{\text{No}}{\text{. of solute}}}}{{{\text{Mass of dissolvable}}\left( {kg} \right)}}$
${\text{Weight of solvent}} = {\text{Weight of solution}} - {\text{Weight of NaCl}}$
Weight of the solvent\[ = 1.0585 \times 1000 - 58.5 = 1Kg\]
Thus the molality
\[Molality{\text{ }} = {\text{ }}\dfrac{1}{1} = 1\]

So, the correct answer is Option B.

Additional note:
Let’s we list out the advantages and disadvantages of using molality as,
The preferred position of utilizing molality as a proportion of focus is that molality is just subject to the mass of solute and the dissolvable which that molality isn't influenced by changes in temperature or weight.
The drawback of molality is, it isn't legitimate in situations where there is no unadulterated substance in a combination. Blends, for example, water and liquor or compounds is a model.

Note: Don't confuse the terms of molality and molarity. The quantity of moles of solute in one liter of arrangement is characterized as the molarity.
The mathematical expression of molarity is,
${\text{Molarity}} = \dfrac{{{\text{Moles of solute}}\left( {Mol} \right)}}{{{\text{Solvent}}\left( L \right)}}$
Keep in mind; molality is utilized to quantify the moles to the mass of the dissolvable and not with the mass of the arrangement.
Example:
First, calculate the number of moles of sodium carbonate using the formula,
Given the molecular weight of sodium carbonate is $106g/mol$.
The mass of sodium carbonate is $25.3g$.
The volume of the solution is $250ml$.
The number of moles of sodium carbonate is,
$Moles = \dfrac{{25.3g}}{{106g/mol}}$
$Moles = 0.239moles$
Now, we estimate the molarity of the solution using the above formula,
$Molarity = \dfrac{{0.239mol}}{{0.25L}}$
$Molarity = 0.956mol/L$
Thus, the molarity of the solution is $0.956mol/L$.