Question

How many grams of NaBr are required to prepare 100 mL of a 0.75M solution?

Answer 1

The weight of NBr required is 1.72 grams.

GIVEN

Volume of the NBr solution = 100 mL

Molarity of the solution = 0.75 M

TO FIND

Weight of NBr

SOLUTION

We can simply solve the above problem as follows,

It is given,

Volume of solution = 100 ml = 100/1000 = 0.1 L

Molarity of the solution = 0.75 M

Let the number of moles of NBr in the solution be m moles

We know that,

Molarity = moles of solute /Volume of solution(in Litres)

Putting the values in the above problem as follows,

0.75 = m/0.1

m = 0.1 × 0.75

m = 0.075 moles

We know that,

Moles = given weight/molecular weight

Putting the values in the above formula we get,

0.075 = given weight /22.99

Given weight = 0.075 × 22.99 = 1.72 grams

Hence, The weight of NBr required is 1.72 grams.

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Answer 2

7.725 g of NaBr is needed to prepare 100 mL of a 0.75 M solution.

Given,

Volume of the solution=100 mL

Molarity of the solution=0.75 M.

To find,

the mass of NaBr needed to prepare 100 mL of a 0.75M solution.

Solution:

• Molarity of a substance in a solution is defined as the moles of that substance in 1 L of solution.
• $Molarity=\frac{Moles}{Volume(L)}$.
• Molarity is a temperature dependent concentration term.

Moles of NaBr present in the solution are:

$Molarity=\frac{Moles}{Volume(L)}\\0.75=\frac{Moles}{\frac{100}{1000} }\\0.75=\frac{Moles}{0.1}\\Moles=0.75X0.1\\Moles=0.075.$

0.075 moles of NaBr are present in the solution.

The molar mass of NaBr is 103 g.

0.075 moles of NaBr will have a mass of:

$Moles=\frac{Mass}{Molar mass} \\0.075=\frac{Mass}{103}\\Mass=0.075X103\\Mass=7.725 g.$

0.075 moles of NaBr will have a mass of 7.725 g.

Thus, the mass of NaBr required is 7.725 g.

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