Question

The weight in grams in KCl (Mol.wt = 74.5) in 100 ml of a 0.1 M KCl solution is:
(A) 74.5
(B) 2024
(C) 0.745
(D) 0.0745

Answer 1

To answer this question we can use the formulas to calculate the number of moles. Appropriate conversion should be done in terms asked in order to avoid arriving at a wrong answer.

Complete step by step answer:
Given,
Molarity = 0.1M
Molecular weight of KCl = 74.5g
Volume of solution = 100mL
We know that,
Molarity and volume can be related by following formula:
Number of moles = Molarity $\times$ Volume of solution ( in liters) …………equation 1
According to the above formula, we have to convert the volume of solution from mL to L.
We know that,
1 liter = 1000 milli-liter
So, 100mL = 0.1L
Number of moles can be represented as 'n'.
n = 0.1M $\times$ 0.1L
= 0.01
There is other definition for mole which may be represented as following equation:
Number of moles = $\dfrac{weight\text{ }in\text{ }g}{Molecular\text{ }weight\text{ }in\text{ }g}$……….equation 2
From the above calculation we found that the number of moles that is n is equal to 0.00001.
Molecular weight of KCl is 74.5g
Now, let's substitute the value of n and the molecular weight of KCl in equation 2 ,
n = $\dfrac{weight\text{ }in\text{ }g}{Molecular\text{ }weight\text{ }in\text{ }g}$
0.00001 = $\dfrac{weight\text{ }in\text{ }g}{74.5g}$
Weight in g = 0.01 $\times$ 74.5
Weight in g = 0.745g
Weight of KCl in grams is 0.745.
The correct answer is option “C” .

Note: Moles can be represented or defined as follows:
-Number of moles of molecules = $\dfrac{weight\text{ }in\text{ }g}{Molecular\text{ }weight\text{ }in\text{ }g}$
- Number of moles of atoms = $\dfrac{weight\text{ }in\text{ }g}{\text{Atomic }weight\text{ }in\text{ }g}$
- Number of moles of gases = $\dfrac{Volume\text{ }in\text{ }NTP}{Standard\text{ }molar\text{ }volume}$
-Number of moles of atoms/molecules/ions/electrons =$\dfrac{Number\text{ }of\text{ }atoms/molecules/ions/electrons}{Avogadro\text{ }constant}$
-Number of moles of solute = Molarity $\times$ Volume of solution ( in liters)
According to our need we can use the appropriate formula and answer this question within no time.