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# How many grams of $C{H_3}OH$ should be added to water to prepare a 150ml solution of 2M $C{H_3}OH$?A. $9.61 \times {10^3}$B .$2.4 \times {10^3}$C. 9.61D. 2.4

Last updated date: 20th Jun 2024
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Hint: Methyl alcohol, $C{H_3}OH$ is primary alcohol and it is a simple aliphatic alcohol. The concentration of $C{H_3}OH$ to be prepared is denoted by molarity here, expressed as M. Molarity of any solution is an indication of number of moles of a solute present in a litre of a solution. It is one of the most commonly used units for expressing the concentration of a solution. The amount of a solute or the volume of a solvent can be determined with the help of the molarity of solution.

Formula used: Here, we will be using two formulae we will be using to calculate the amount of $C{H_3}OH$ required to add in water to make a 150ml solution of 2M $C{H_3}OH$.
$n = M \times V$
Here, $n =$ number of moles of $C{H_3}OH$
$M =$ Molarity of solution of water and $C{H_3}OH$
$V =$ Volume of solution
$Mass \;of\; C{H_3}OH \;required = n \times mola\;rmass$

Generally, alcohols are soluble in water. The first aliphatic three alcohols namely, methanol, ethanol and propanol are completely soluble in water. This is possible due to formation of hydrogen bonds between the hydroxyl groups in alcohols with water molecules.
Here we have to calculate the amount of solute that is methyl alcohol required to prepare a 150ml solution of 2M in water. Let us now note down the quantities given:
Molarity of solution (M) = 2M
Volume of the solution (V) = 150ml
Step 1: To calculate the amount of $C{H_3}OH$ required to prepare a 150ml solution of $C{H_3}OH$ in water of 2M concentration. For that, we need to first calculate the moles of $C{H_3}OH$ required, which will help us to determine the amount of $C{H_3}OH$ needed.
The molarity of a solution can also be described as the number of moles of solute divided by the volume of the solution in litres. The formula of molarity is given below.
$M = \dfrac{n}{V}$
Here, n is the number of moles of solute, M is the molarity of solution and V is the volume of solution in litres.
We have values for both molarity of the solution and volume of solution, so we need to rearrange the above formula to obtain the number of moles of solute.
On rearranging the formula, we get
$n = M \times V$……………..(1)
Now, we require volume in litres but it is given in millilitres. Let us convert the volume into litres by dividing the volume by 1000.
Volume of the solution=$\dfrac{{150}}{{1000}}L = 0.15L$
Let us substitute respective values in formula (1), we get,
$n = 2 \times 0.15$
$\therefore n = 0.3mol$
The number of moles of $C{H_3}OH$ required is 0.3mol.
Step 2: Now, we know the number of moles of $C{H_3}OH$ required, from that we can calculate the amount of $C{H_3}OH$ required.
The number of moles can also be calculated using the following formula.
$n = \dfrac{m}{{molar\,mass}}$……………..(2)
If we rearrange the formula (2), we will be able to calculate the amount of $C{H_3}OH$ required in grams.
Here, n is the number of moles of a compound and m is the mass of that compound in grams.
On rearranging the formula (2), we get,
$m = n \times molar\;mass$
The molar mass of $C{H_3}OH = (1 \times 12) + (3 \times 1) + (1 \times 16) + (1 \times 1) = 32$
The molar mass of $C{H_3}OH$ is 32g/mol.
Now, on putting the corresponding values in formula (2), we get,
$m = 0.3mol \times 32g/mol$
$\therefore m = 9.61g$
To prepare 150ml solution 2M $C{H_3}OH$, we need to add 9.61g of $C{H_3}OH$ in water.

So, the correct answer is C.

Note: While calculating the amount of $C{H_3}OH$ required in grams, students should keep in mind that volume needs to be in litres.
Also, keep in mind that the term ‘volume’ is referring to that of solution, not solvent. This means that we need to consider the volume of solution of water and $C{H_3}OH$.