Answer to Question #314039 in General Chemistry for Sai

Question

1) What is the molarity of 1.28 mol of CA(NO3)2 dissolved in 1.50 L of solution?

2) How many moles are contained in 0.210M KOH?

Answer 1

1)Answer= 0.8533 M

Explanation:

Molarity of a solution is defined as the number of moles of the Solute present in 1000 ml (or) 1-Litre of the solution.

Here 1.28 moles of Ca (NO3)2 is present in 1.5 Litres of the solution.

So 1Litre of the solution contains =1.28 /1.5 = 0.8533 M

2)Let's supposed it's dissolved in water. 0.210M means there are 0.210 moles of KOH dissolved in 1 litre of water, which equals to ca. 1 kg of water (slightly less in fact). Water has molar mass of ca. 18g/mol (2×M(H)+M(O)=2×1+16), therefore, if we have 1kg of it, it means we have n=m/M(H2O)=1000g/(18g/mol)=55,56mol of water.