Question

The weight in grams of $KCl\;(Mol.Wt. = 74.5)$ in $100ml$ of $0.1\;M\;KCl$ solution.
A. $74.5$
B. $2024$
C. $0.745$
D. $0.0745$

Answer 1

Here we know that, the molarity that is the concentration of solute in the solution is given, the weight of solute taken by weighing in the weighing pan needs to be determined. The molecular weight of the potassium chloride is constant and is given. It can be used to calculate the molar mass which is equal to weight weighted in a weighing pan per molecular weight.

Complete step by step solution
We know that the number of moles present in the solution can be calculated by dividing the 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. One mole of solute is the weight of solute per molecular weight of solute in grams.
The formula of molarity can be used to find the weight of the potassium chloride. The formula is as shown below in the calculation.
Molecular weight of solute is the sum weight of all the atoms present in the molecule. Now, let us do our calculation as follows;

$$Molarity = \dfrac{{moles\;of\;solute}}{{volume\;of\;solution\;(liter)}}\\\ Moles\;of\;solute = \dfrac{{weight}}{{molecular\;weight}}\\\ Molarity = \dfrac{{weight}}{{Mwt}} \times \dfrac{{1000}}{{V(ml)}}$$

Let’s substitute the known values in the above formula.

$$0.1 = \dfrac{{weight}}{{74.5}} \times \dfrac{{1000}}{{100}}\\\ weight = 0.745\;ml$$

Hence, the weight $= \;0.745ml$ of $KCl\;(Mol.Wt. = 74.5)$ in $100ml$ of $0.1\;M\;KCl$ solution.

The option is (C) $0.745ml$, which is our answer.

The concentration of the solution can be found by using these formulas in the laboratory for doing experiments.

Note:
Do not get confused between the word molality and molarity as the amount of solute in moles dissolved in one kilogram of solvent. The formula of molality is as follows:
$Molality = \dfrac{{moles\;of\;solute}}{{kilogram\;of\;solvent}}$