Answer:
We need 44.63 g of KBr to make 250 mL of a 1.50 M KBr solution.
Explanation:
To solve this problem it is important to remember the definition of Molarity.
Molarity is a concentration unit that represents the moles of solute that are contained in 1 L (or 1000 mL) of solution.
Let's solve the problem by using the Rule of three, this helps us to reasoning the problem instead of using the formula of molarity.
1st) Find the number of moles that are containded in the 1.5 M concentration following the Molarity definition:
1.50 M means that we have 1.50 moles of solute in 1000 mL of solution.
So, if 1000 mL of solution has 1.50 moles of solute, the 250 mL we need to make will contain x moles. The rule of three is:
1000 mL -------------- 1.50 moles
250 mL --------------- x = (250 mL . 1.50 moles)/1000 mL = 0.375 moles
2nd) Find the Molecular weight (Molar mass) of the solute KBr:
To do this, look for the atomic weight of each atom and sum them.
K atomic weight + Br atomic weight = KBr molecular weight
39 g/mol + 80 g/mol = 119 g/mol
The molecular weight means that 1 mol of KBr weights 119g.
3rd) Calculate the mass of KBr:
We can calculate this with another Rule of three.
If 1 mol of KBr weights 119 g, the 0.375 moles we need to prepare the solution weights x. The calculation is:
1 mol ------------ 119 g
0.375 moles ------------ x = (0.375 moles . 119 g)/1 mol = 44.63g of KBr
We have that the mass of KBr is mathematically given as
x= 44.63g of KBr
Generally, we find the Molecular weight (Molar mass) of the solute
KBr, K atomic weight + Br atomic weight = KBr molecular weight
39 g/mol + 80 g/mol = 119 g/mol
The molecular weight means that
1 mol of KBr weights 119g.
If 1 mol of KBr weights 119 g, then 0.375 moles we need to prepare the solution weights x.
1 mol =119 g
x= (0.375 moles . 119 g)/1 mol
x= 44.63g of KBr
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