Question

What is the molarity of a solution made by dissolving 65.0 g of NaNO3 in enough water to make 0.350 L of solution?

Answer 1

First we need the molar mass of NaNO_3:-

`\\ \sf\longmapsto 23+14+3(16)`

`\\ \sf\longmapsto 37+48`

`\\ \sf\longmapsto 85g/mol`

Now we need moles of NaNO_3

`\\ \sf\longmapsto No\:of\:moles=(Given\:mass)/(Molar\:mass)`

`\\ \sf\longmapsto No\:of\:moles=(65)/(85)`

`\\ \sf\longmapsto No\:of\:moles=0.765mol`

Now

`\\ \sf\longmapsto Molarity=(Moles\:of\:solute)/(Vol\:of\:soln.\:in\:L)`

`\\ \sf\longmapsto Molarity=(0.765)/(0.350)`

`\\ \sf\longmapsto Molarity=2.2M`