How many grams of KMnO4 are required to prepare 1.0 L of a solution of 1.5 M KMNO4?
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Molarity=No of moles of solute(KMnO₄)/Volume of solution(L)
Given values:
Molarity=1.5
Volume of solution=1 L
Lets find No of moles of KMnO₄:
1.5= No of moles of KMnO₄/1
No of moles=1.5
Moles=given mass/Molar mass
Molar mass of KMNO₄=158
Here we need to find the given mass
1.5=given mass/158
given mass=1.5×158
=237g
Hence we require 237 g of KMnO₄ to make a solution of 1.5 M
Given values:
Molarity=1.5
Volume of solution=1 L
Lets find No of moles of KMnO₄:
1.5= No of moles of KMnO₄/1
No of moles=1.5
Moles=given mass/Molar mass
Molar mass of KMNO₄=158
Here we need to find the given mass
1.5=given mass/158
given mass=1.5×158
=237g
Hence we require 237 g of KMnO₄ to make a solution of 1.5 M
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