surendar selvam answered 1 litre @ 0.50 mol / litre has 0.50 moles of soluteNiCl2 from the salt NiCl2·6H2O whose Formula Weight is 237.71g/mol0.50 moles @ 237.71g/mol = 118.855 gramshow would you prepare 1.00 L of a 0.50 M ?.... Even though there arefew significant digits requested , I would still weigh out 118.8 gramsofNiCl2·6H2O , which I would dissolve into some water, & once dissolved I would dilute it up to 1.00 litreswhen I weigh out 0.50 moles of NiCl2·6H2O , it gives me 0.5 moles of NiCl2 with a little excess baggage in the form of 6H2O======Sodium carbonate (with a molar mass of 105.99 g/mol)from the pure solid0.50 moles @ 105.99 g/ mol = 52.995 gramshow would you prepare 1.00 L of a 0.50 M ?.... Even though thereare few significant digits requested , I would still weigh out 52.0grams of Na2CO3 , which I would dissolve into some water, & oncedissolved I would dilute it up to 1.00 litres
